NCERT Solutions Class 11 Mathematics Chapter 16

Q.1 In each of the following Exercises 1 to 7, describe the sample space for the indicated experiment.

1. A coin is tossed three times.
2. A die is thrown two times.
3. A coin is tossed four times.
4. A coin is tossed and a die is thrown.
5. A coin is tossed and then a die is rolled only in case a head is shown on the coin.
6. 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
7. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.

Ans

$\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$

3. Sample space for a coin tossed four times is:

$\left\{\begin{array}{l}\mathrm{HHHH},\mathrm{HHHT},\mathrm{HHTH},\mathrm{HTHH},\mathrm{THHH},\mathrm{HHTT},\mathrm{HTTH},\mathrm{HTHT},\\ \mathrm{THHT},\mathrm{THTH},\mathrm{TTHH},\mathrm{TTTH},\mathrm{HTTT},\mathrm{TTHT},\mathrm{THTT},\mathrm{TTTT}\end{array}\right\}$

4. Sample space for a tossed coin thrown die is:

$\{\mathrm{H}1,\mathrm{H}2,\mathrm{H}3,\mathrm{H}4,\mathrm{H}5,\mathrm{H}6,\mathrm{T}1,\mathrm{T}2,\mathrm{T}3,\mathrm{T}4,\mathrm{T}5,\mathrm{T}6\}$

5. Sample space when a coin is tossed and then a die is rolled only when head is shown on the coin:

$\{\mathrm{H}1,\mathrm{H}2,\mathrm{H}3,\mathrm{H}4,\mathrm{H}5,\mathrm{H}6,\mathrm{T}\}$

6. The sample space for the experiment in which a room is selected and then a person is:

$\{{\mathrm{XB}}_1,{\mathrm{XB}}_2,{\mathrm{XG}}_1,{\mathrm{XG}}_2,{\mathrm{YB}}_3,{\mathrm{YG}}_3,{\mathrm{YG}}_4,{\mathrm{YG}}_5\}$

7. Dice in red colour, white colour and blue colour are shown by R, W and B respectively. Then the required sample space is

$\{\mathrm{R}1,\mathrm{R}2,\mathrm{R}3,\mathrm{R}4,\mathrm{R}5,\mathrm{R}6,\mathrm{W}1,\mathrm{W}2,\mathrm{W}3,\mathrm{W}4,\mathrm{W}5,\mathrm{W}6,\mathrm{B}1,\mathrm{B}2,\mathrm{B}3,\mathrm{B}4,\mathrm{B}5,\mathrm{B}6\}$

Q.2 An experiment consists of recording boy–girl composition of families with 2 children.

(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
(ii) What is the sample space if we are interested in the number of girls in the family?

Ans

The boy and girl are represented by B and G respectively.
(i) The required sample space is:
{ BB,BG,GB,GG }

(ii) Sample space according to number of children in family is:
{ 0, 1, 2 }

Q.3 A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.

Ans

A red ball and white ball are represented by R and W respectively.
The required sample space is:
{ RW, WR, WW }.

Q.4 An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.

Ans

The head and tail are represented by H and T respectively.
The required sample space, when coin and die is thrown according to given condition, is:
{ HH, HT, T1, T2, T3, T4, T5, T6 }

Q.5 Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non – defective(N). Write the sample space of this experiment.

Ans

The required sample space is:
{DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}

Q.6 A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?

Ans

Sample space according to given conditions is:
{ T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66 }

Q.7 The numbers 1, 2, 3 and 4 are written separatly on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.

Ans

The required sample space is:
{(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

Q.8 An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.

Ans

The required sample space is:
{1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T }

Q.9 A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment.

Ans

The required sample space is:
{TR₁, TR₂, TB₁, TB₂, TB₃, H1, H2, H3, H4, H5, H6}

Q.10 A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Ans

Sample space when a die is thrown until a six is obtained.
{ 6, (1, 6), (2,6), (3,6), (4,6), (5,6), (1, 1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2,1,6), (2, 2, 6),…, (2,5,6), (3,1,6), (3, 2, 6), (3,3,6), (3, 4, 6), (3, 5, 6) …}

Q.11 A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

Ans

Here, E = {4}, F = {2, 4, 6}
E ∩ F = {4}
Since, E ∩ F ≠

$\mathrm{\varphi }$

So, E and F are not mutually exclusive events.

Q.12 A die is thrown. Describe the following events:

(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3

Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′

Ans

Here,
(i) A = {1, 2, 3, 4, 5, 6}
(ii) B = {} = Φ
(iii) C = {3, 6}
(iv) D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6},
A ∩ B = {} = Φ
B ∪ C = {3, 6}
E ∩ F = {6}
D ∩ E = {} = Φ
A – C = {1, 2, 4, 5}
D – E = {1, 2, 3}
E ∩ F’ = {6} ∩ [{1, 2, 3, 4, 5, 6} – {3, 4, 5, 6}]
= {6} ∩ {1, 2}
= {} = Φ
F’ = {1, 2}

Q.13 An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

A: the sum is greater than 8,
B: 2 occurs on either die
C: the sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?

Ans

$\mathrm{S}=\left\{\begin{array}{l}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)\\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)\\ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)\\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)\\ (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)\\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\end{array}\right\}$

A = {(3, 6), (4, 5), (4, 6),(5, 4), (5, 5),(5, 6), (6, 3),(6, 4), (6, 5),(6, 6)}
B = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}
C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Since, A ∩ B =

$\mathrm{\varphi }$

. So, A and B are mutually exclusive.
And B ∩ C =

$\mathrm{\varphi }$

. So, B and C are mutually exclusive.

Q.14 Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”.
Which events are (i) mutually exclusive? (ii) simple? (iii) Compound?

Ans

$\begin{array}{l}\mathrm{S}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right),\left(\mathrm{TTT}\right)\}\\ \mathrm{A}=\left\{\left(\mathrm{HHH}\right)\right\}\\ \mathrm{B}=\{\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right)\}\\ \mathrm{C}=\left\{\left(\mathrm{TTT}\right)\right\}\\ \mathrm{D}=\{\left(\mathrm{HTT}\right),\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{HHT}\right)\}\\ \left(\mathrm{i}\right)\mathrm{Since},\text{ A}\cap \text{B =}\mathrm{\varphi },\text{so, A and B are Mutually exclusive.}\\ \mathrm{Since},\text{ A}\cap \text{C =}\mathrm{\varphi }\text{so, A and C are Mutually exclusive.}\\ \mathrm{Since},\text{ B}\cap \text{C =}\mathrm{\varphi }\text{so, B and C are Mutually exclusive.}\\ \mathrm{Since},\text{ C}\cap \text{D =}\mathrm{\varphi }\text{so, C and D are Mutually exclusive.}\\ \left(\mathrm{ii}\right)\text{A and C are simple events because they have single element.}\\ \left(\mathrm{iii}\right)\mathrm{Compound}\text{events are B and D.}\end{array}$

Q.15 Three coins are tossed. Describe

(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.

Ans

$\begin{array}{l}\mathrm{S}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right),\left(\mathrm{TTT}\right)\}\\ \left(\mathrm{i}\right)\text{Two mutually exclusive events are:}\\ \left\{\left(\mathrm{HHH}\right)\right\}\text{and}\left\{\left(\mathrm{TTT}\right)\right\}.\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Three events which are mutually exclusive and\hspace{0.17em}}\\ \text{exhaustive:\hspace{0.17em}}\\ \text{A}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right)\}\\ \mathrm{B}=\{\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right)\}\mathrm{and}\\ \mathrm{C}=\{\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right),\left(\mathrm{TTT}\right)\}\\ \because \mathrm{A}\cap \mathrm{B}=\mathrm{\varphi },\mathrm{B}\cap \mathrm{C}=\mathrm{\varphi },\mathrm{A}\cap \mathrm{C}=\mathrm{\varphi }\text{and A}\cup \text{B}\cup \mathrm{C}=\text{S}\\ \left(\mathrm{iii}\right)\mathrm{D}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right)\}\\ \mathrm{and}\mathrm{E}=\{\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right)\}\\ \because \mathrm{D}\cap \mathrm{E}\ne \mathrm{\varphi },\text{so, D and E are not mutually exclusive.}\\ \left(\mathrm{iv}\right)\mathrm{F}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\}\\ \mathrm{G}=\left\{\left(\mathrm{TTH}\right)\right\}\\ \mathrm{Since},\mathrm{F}\cap \mathrm{G}=\mathrm{\varphi }\text{but F}\cup \text{G}=\text{S}\\ \text{So, F and G are exclusive but not exhaustive.}\\ \left(\mathrm{v}\right)\text{H}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right)\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}K}=\{\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right)\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}L}=\{\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right)\}\\ \because \mathrm{H}\cap \mathrm{K}=\mathrm{\varphi },\mathrm{H}\cap \mathrm{L}=\mathrm{\varphi }\\ \mathrm{K}\cap \mathrm{L}=\mathrm{\varphi }\text{and H}\cup \text{K}\cup \mathrm{L}\ne \text{S}\\ \text{So, H, K and L are mutually exclusive but not exhaustive.}\end{array}$

Q.16 Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5.

Describe the events
(i) A′ (ii) not B ,
(iii) A or B (iv) A and B
(v) A but not C (vi) B or C
(vii) B and C (viii) A ∩ B′∩C′

Ans

$\begin{array}{l}\mathrm{S}=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \text{A}=\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \text{B}=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\end{array}\right\}\\ \mathrm{C}=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(2,1),\\ (2,2),(2,3),(3,1),(3,2),(4,1)\end{array}\right\}\end{array}$ $\begin{array}{l}\left(\mathrm{i}\right)\text{A’}=\mathrm{S}-\mathrm{A}\\ =\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\end{array}\right\}=\mathrm{B}\\ \left(\mathrm{ii}\right)\text{not B}=\mathrm{S}-\mathrm{B}\\ =\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}=\mathrm{A}\\ \left(\mathrm{iii}\right)\mathrm{A}\text{or B}=\mathrm{A}\cup \mathrm{B}\\ =\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}=\mathrm{S}\\ \left(\mathrm{iv}\right)\text{A and B}=\mathrm{A}\cap \mathrm{B}=\mathrm{\varphi }\\ \left(\mathrm{v}\right)\mathrm{A}\text{but not C}=\mathrm{A}-\mathrm{C}\\ =\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ -\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(2,1),\\ (2,2),(2,3),(3,1),(3,2),(4,1)\end{array}\right\}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\left\{\begin{array}{l}(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),\\ (4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \left(\mathrm{vi}\right)\text{B or C}=\mathrm{B}\cup \mathrm{C}\\ =\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),\\ (2,2),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),\\ (3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),\\ (5,6)\end{array}\right\}\\ \left(\mathrm{vii}\right)\mathrm{B}\mathrm{and}\mathrm{C}=\mathrm{B}\cap \mathrm{C}\\ =\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}\\ \left(\mathrm{viii}\right)\mathrm{A}\cap \mathrm{B}‘=\mathrm{A}\cap (\mathrm{S}-\mathrm{B})\\ =\mathrm{A}\cap \mathrm{A}\left[\mathrm{From}\left(\mathrm{ii}\right)\right]\\ =\mathrm{A}\\ =\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \mathrm{C}‘=(\mathrm{S}-\mathrm{C})\\ =\left\{\begin{array}{l}(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),\\ (3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),\\ (6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \mathrm{A}\cap \mathrm{B}‘\cap \mathrm{C}‘=\left\{\begin{array}{l}(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),\\ (4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\end{array}$

Q.17 Refer to question 6 above, state true or false: (Give reason for your answer)

(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) A = B′
(iv) A and C are mutually exclusive
(v) A and B′ are mutually exclusive.
(vi) A′, B′, C are mutually exclusive and exhaustive.

Ans

1. True, because A ∩ B = Φ
2. True, because A ∩ B = Φ and A È B = S.
3. True, because B’= S – B = A.
4. False, because A ∩ C ≠ Φ
5. False, because A ∩ B’ = A ∩ A ≠ Φ [Here, B’ = A]
6. False, because A’ ∩ B’ = Φ, B’ ∩ C ≠ Φ

Q.18

$\begin{array}{l}\mathbf{\text{Which of the following cannot be valid assignment of probabilities}}\\ \mathbf{\text{for outcomes of sample Space S = {}}{\mathbf{\omega }}_{\mathbf{1}}\mathbf{\text{,}}{\mathbf{\omega }}_{\mathbf{2}}\mathbf{\text{,}}{\mathbf{\omega }}_{\mathbf{3}}\mathbf{\text{,}}{\mathbf{\omega }}_{\mathbf{4}}\mathbf{\text{,}}{\mathbf{\omega }}_{\mathbf{5}}\mathbf{\text{,}}{\mathbf{\omega }}_{\mathbf{6}}\mathbf{\text{,}}{\mathbf{\omega }}_{\mathbf{7}}\mathbf{\text{}}}\\ \begin{array}{cccccccc}\mathbf{Assignment}& {\mathbf{\omega }}_{\mathbf{1}}& {\mathbf{\omega }}_{\mathbf{2}}& {\mathbf{\omega }}_{\mathbf{3}}& {\mathbf{\omega }}_{\mathbf{4}}& {\mathbf{\omega }}_{\mathbf{5}}& {\mathbf{\omega }}_{\mathbf{6}}& {\mathbf{\omega }}_{\mathbf{7}}\\ \mathbf{\left(}\mathbf{a}\mathbf{\right)}& \mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.01}& \mathbf{0}\mathbf{.05}& \mathbf{0}\mathbf{.03}& \mathbf{0}\mathbf{.01}& \mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.6}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}& \frac{\mathbf{1}}{\mathbf{7}}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}& \mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.3}& \mathbf{0}\mathbf{.4}& \mathbf{0}\mathbf{.5}& \mathbf{0}\mathbf{.6}& \mathbf{0}\mathbf{.7}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}& \mathbf{-}\mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.3}& \mathbf{0}\mathbf{.4}& \mathbf{-}\mathbf{0}\mathbf{.2}& \mathbf{0}\mathbf{.1}& \mathbf{0}\mathbf{.3}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}& \frac{\mathbf{1}}{\mathbf{14}}& \frac{\mathbf{2}}{\mathbf{14}}& \frac{\mathbf{3}}{\mathbf{14}}& \frac{\mathbf{4}}{\mathbf{14}}& \frac{\mathbf{5}}{\mathbf{14}}& \frac{\mathbf{6}}{\mathbf{14}}& \frac{\mathbf{15}}{\mathbf{14}}\end{array}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right){\text{ω}}_{\text{1}}{\text{+ ω}}_{\text{2}}{\text{+ ω}}_{\text{3}}{\text{+ ω}}_{\text{4}}{\text{+ ω}}_{\text{5}}{\text{+ ω}}_{\text{6}}{\text{+ ω}}_{\text{7}}\\ \text{= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6}\\ \text{= 1}\\ \text{Yes, it is a valid assignment of probabilities.}\\ \left(\text{b}\right){\text{ω}}_{\text{1}}{\text{+ ω}}_{\text{2}}{\text{+ ω}}_{\text{3}}{\text{+ ω}}_{\text{4}}{\text{+ ω}}_{\text{5}}{\text{+ ω}}_{\text{6}}{\text{+ ω}}_{\text{7}}\\ \text{=}\frac{\text{1}}{\text{7}}\text{+}\frac{\text{1}}{\text{7}}\text{+}\frac{\text{1}}{\text{7}}\text{+}\frac{\text{1}}{\text{7}}\text{+}\frac{\text{1}}{\text{7}}\text{+}\frac{\text{1}}{\text{7}}\text{+}\frac{\text{1}}{\text{7}}\\ \text{=}\frac{\text{7}}{\text{7}}\\ \text{= 1}\\ \text{Yes, it is a valid assignment of probabilities.}\\ \left(\text{c}\right){\text{ω}}_{\text{1}}{\text{+ ω}}_{\text{2}}{\text{+ ω}}_{\text{3}}{\text{+ ω}}_{\text{4}}{\text{+ ω}}_{\text{5}}{\text{+ ω}}_{\text{6}}{\text{+ ω}}_{\text{7}}\\ \text{= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7}\\ \text{= 2.8}\\ \text{No, it is not a valid assignment of probabilities.}\\ \text{Because sum of probabilities is greater than 1.}\\ \left(\text{d}\right){\text{ω}}_{\text{1}}{\text{+ ω}}_{\text{2}}{\text{+ ω}}_{\text{3}}{\text{+ ω}}_{\text{4}}{\text{+ ω}}_{\text{5}}{\text{+ ω}}_{\text{6}}{\text{+ ω}}_{\text{7}}\\ \text{= -\hspace{0.17em}0.1 + 0.2 + 0.3 + 0.4 +}\left(\text{– 0.2}\right)\text{+ \hspace{0.17em}0.1 + 0.3}\\ \text{No, it is not a valid assignment of probabilities.}\\ \text{Because probability can not be negative.}\\ \left(\text{e}\right){\text{ω}}_{\text{1}}{\text{+ ω}}_{\text{2}}{\text{+ ω}}_{\text{3}}{\text{+ ω}}_{\text{4}}{\text{+ ω}}_{\text{5}}{\text{+ ω}}_{\text{6}}{\text{+ ω}}_{\text{7}}\\ \text{=}\frac{\text{1}}{\text{14}}\text{+ \hspace{0.17em}}\frac{\text{2}}{\text{14}}\text{+}\frac{\text{3}}{\text{14}}\text{+}\frac{\text{4}}{\text{14}}\text{+}\frac{\text{5}}{\text{14}}\text{+}\frac{\text{6}}{\text{14}}\text{+}\frac{\text{15}}{\text{14}}\\ \text{=}\frac{\text{1 + 2 + 3 + 4 + 5 + 6 + 15}}{\text{14}}\end{array}$

Q.19 A coin is tossed twice, what is the probability that at least one tail occurs?

Ans

Sample space when a coin is tossed twice,
S(E) = {HH, HT, TH, TT}
Favorable events of getting at least one tail
E = {HT, TH, TT}

$\begin{array}{l}\mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{S}\left(\mathrm{E}\right)}\\ =\frac{3}{4}\\ \mathrm{Thus},\text{the probability of at least one tail is}\frac{3}{4}.\end{array}$

Q.20 A die is thrown; find the probability of following events:

(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.

Ans

Sample space of a thrown die is:
S(E) = {1,2,3,4,5,6}
(i) Prime number (E) = {2, 3, 5}

$\mathrm{P}\left(\text{Prime number}\right)=\frac{3}{6}=\frac{1}{2}$

(ii) Events of getting a number greater than or equal to 3
= {3, 4, 5, 6}

$\begin{array}{l}\mathrm{P}\left(\text{E}\right)=\frac{4}{6}=\frac{2}{3}\\ \mathrm{Thus},\text{the required probability is}\frac{2}{3}.\end{array}$

(iii) Event of getting a number less than or equal to one
= {1}

$\begin{array}{l}\mathrm{P}\left(\text{E}\right)=\frac{1}{6}\\ \mathrm{Thus},\text{the required probability is}\frac{1}{3}.\end{array}$

(iv) Event of getting a number more than 6
= {}

$\begin{array}{l}\mathrm{P}\left(\text{E}\right)=\frac{0}{6}=0\\ \mathrm{Thus},\text{the required probability is}0.\end{array}$

(v) Event of getting a number less than 6
= {1, 2, 3, 4, 5}

$\begin{array}{l}\mathrm{P}\left(\text{E}\right)=\frac{5}{6}\\ \mathrm{Thus},\text{the required probability is}\frac{5}{6}.\end{array}$

Q.21 A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace (ii) black card.

Ans

(a) There are 52 points in the sample space.
(b) Number of ace of spades = 1

$\begin{array}{l}\mathrm{P}\left(\mathrm{ace}\text{of spades}\right)=\frac{1}{52}\\ \mathrm{Thus},\text{the probability of an ace of spades is}\frac{1}{52}\text{.}\end{array}$

(c) (i) Number of aces in a deck of 52 cards = 4

$\begin{array}{l}\mathrm{P}\left(\mathrm{ace}\right)=\frac{4}{52}=\frac{1}{13}\\ \mathrm{Thus},\text{the probability of an ace card is}\frac{1}{13}.\end{array}$

(ii) Number of black cards in a deck of 52 cards

$\begin{array}{l}\mathrm{P}\left(\mathrm{ace}\right)=\frac{26}{52}=\frac{1}{2}\\ \mathrm{Thus},\text{the probability of an ace card is}\frac{1}{2}.\end{array}$

Q.22 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed, find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Ans

$\begin{array}{l}\mathrm{Sample}\text{events when one coin}\left(\mathrm{faced}\text{1 and 6}\right)\text{and a dice:}\\ \text{S}=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \mathrm{n}\left(\mathrm{S}\right)=12\\ \left(\mathrm{i}\right)\mathrm{Events}\text{in which sum is 3:}\\ \text{E}=\left\{(1,2)\right\}\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{1}{12}\end{array}$ $\begin{array}{l}\mathrm{Thus}\text{, the required probability is}\frac{1}{12}.\\ \left(\mathrm{ii}\right)\mathrm{Events}\text{in which sum is 12:}\\ \text{E}=\left\{(6,6)\right\}\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{1}{12}\\ \mathrm{Thus}\text{, the required probability is}\frac{1}{12}.\end{array}$

Q.23 There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Ans

Number of people in a council = 10
Number of women in a council = 6

$\begin{array}{l}\mathrm{P}\left(\mathrm{woman}\right)=\frac{6}{10}=\frac{3}{5}\\ \mathrm{Thus}\text{, the required probability is}\frac{3}{5}.\end{array}$

Q.24 A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Ans

Sample space when a coin is tossed four times:
{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT,THHT, THTH, TTHH, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

A person gain on getting each head = ₹ 1
A person loses on getting each tail = ₹ 1.50
Gain on getting HHHH = ₹ 4
Gain on getting HHHT = ₹ 1.50
Gain on getting HHTH = ₹ 1.50
Gain on getting HTHH = ₹ 1.50
Gain on getting THHH = ₹ 1.50
Loss on getting HHTT = ₹ 1.00
Loss on getting HTHT = ₹ 1.00
Loss on getting THHT = ₹ 1.00
Loss on getting THTH = ₹ 1.00
Loss on getting TTHH = ₹ 1.00
Loss on getting HTTH = ₹ 1.00
Loss on getting TTTH = ₹ 3.50
Loss on getting TTHT = ₹ 3.50
Loss on getting THTT = ₹ 3.50
Loss on getting HTTT = ₹ 3.50
Loss on getting TTTT = Rs. 6

$\begin{array}{c}\mathrm{P}\left(\mathrm{Winning}\mathrm{₹}4\right)=\frac{1}{16}\\ \mathrm{P}(\mathrm{Winning}\mathrm{₹}1.50)=\frac{4}{16}\\ =\frac{1}{4}\\ \mathrm{P}(\mathrm{Losing}\mathrm{₹}1.00)=\frac{6}{16}\\ =\frac{3}{8}\\ \mathrm{P}(\mathrm{Losing}\mathrm{₹}3.50)=\frac{4}{16}\\ =\frac{1}{4}\end{array}$ $\mathrm{P}(\mathrm{Losing}\mathrm{₹}6.00)=\frac{1}{16}$

Q.25 Three coins are tossed once. Find the probability of getting

(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head (vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails

Ans

Sample space of tossing three coins is:
{HHH, HHT, HTH, THH, HTT, THT, HTT, TTT}

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}P}\left(3\text{heads}\right)=\frac{1}{8}\\ \left(\mathrm{ii}\right)\text{\hspace{0.33em}P}\left(2\text{heads}\right)=\frac{3}{8}\\ \left(\mathrm{iii}\right)\text{\hspace{0.33em}P}\left(\mathrm{at}\text{least 2 heads}\right)=\frac{4}{8}=\frac{1}{2}\\ \left(\mathrm{iv}\right)\text{\hspace{0.33em}P}\left(\mathrm{at}\mathrm{most}2\text{heads}\right)=\frac{7}{8}\\ \left(\text{v}\right)\text{\hspace{0.33em}P}\left(\mathrm{No}\text{head}\right)=\frac{1}{8}\\ \left(\mathrm{vi}\right)\text{\hspace{0.33em}P}\left(3\text{tails}\right)=\frac{1}{8}\\ \left(\mathrm{vii}\right)\text{\hspace{0.33em}P}\left(\mathrm{exactly}\text{2 tails}\right)=\frac{3}{8}\\ \left(\mathrm{viii}\right)\text{\hspace{0.33em}P}\left(\mathrm{No}\text{tail}\right)=\frac{1}{8}\\ \left(\mathrm{ix}\right)\text{\hspace{0.33em}P}\left(\mathrm{at}\mathrm{most}2\text{tails}\right)=\frac{7}{8}\end{array}$

Q.26

$\mathbf{\text{If }}\frac{\mathbf{\text{2}}}{\mathbf{\text{11}}}\mathbf{\text{is the probability of an event, what is the probability of the event ‘not A’.}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{probability of an event be}\mathrm{P}\left(\mathrm{A}\right).\mathrm{Then},\\ \mathrm{P}\left(\mathrm{A}\right)=\frac{2}{11}\\ \mathrm{P}\left(\mathrm{not}\mathrm{A}\right)=1-\mathrm{P}\left(\mathrm{A}\right)\\ =1-\frac{2}{11}\\ =\frac{11-2}{11}\\ =\frac{9}{11}\\ \mathrm{Thus},\text{the probability of the event ‘not A’ is}\frac{9}{11}.\end{array}$

Q.27 A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel (ii) a consonant

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in the word “}\mathrm{ASSASSINATION}“=13\\ \mathrm{Number}\text{of vowels in the word “}\mathrm{ASSASSINATION}“=6\\ \mathrm{Number}\text{of consonants in the word “}\mathrm{ASSASSINATION}“=13-6\\ =7\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{a}\text{vowel}\right)=\frac{6}{13}[\because \mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{S}\left(\mathrm{E}\right)}]\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{a}\text{consonant}\right)=\frac{7}{13}\end{array}$

Q.28 In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

Ans

Total numbers from 1 to 20 = 20
Numbers to be selected = 6

$\begin{array}{l}\mathrm{Number}\text{of ways to select 6 numbers from 20 numbers}\\ ={}^{20}{\mathrm{C}}_6\\ =\frac{20!}{6!(20-6)!}\\ =\frac{20\times 19\times 18\times 17\times 16\times 15\times 14!}{6\times 5\times 4\times 3\times 2\times 1\times 14!}\\ =38760\\ \mathrm{Number}\text{of ways to fix 6 numbers for winning lottery = 1}\\ \mathrm{P}\left(\mathrm{Winning}\text{the prize}\right)=\frac{1}{38760}\end{array}$

Q.29 Check whether the following probabilities P(A) and P(B) are consistently defined

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

Ans

$\begin{array}{l}\left(\text{i}\right)\mathrm{Here},\text{\hspace{0.17em}P}(\mathrm{A}\cap \mathrm{B})>\mathrm{P}\left(\mathrm{A}\right)[\because \text{P}(\mathrm{A}\cap \mathrm{B})=0.6\text{and P}\left(\mathrm{A}\right)=0.5]\\ \mathrm{So},\text{the probabilities}\mathrm{}\text{P}\left(\text{A}\right)\text{and P}\left(\text{B}\right)\text{are not consistently defined.}\\ \left(\mathrm{ii}\right)\mathrm{Here},\mathrm{P}(\mathrm{A}\cup \mathrm{B})>\mathrm{P}\left(\mathrm{A}\right)\text{and}\mathrm{P}(\mathrm{A}\cup \mathrm{B})>\mathrm{P}\left(\mathrm{B}\right).\\ \mathrm{So},\text{the probabilities}\mathrm{}\text{P}\left(\text{A}\right)\text{and P}\left(\text{B}\right)\text{are consistently defined.}\end{array}$

Q.30

$\begin{array}{l}\mathbf{\text{Fill in the blanks in following table:}}\\ \begin{array}{ccccc}& \mathbf{\text{P(A)}}& \mathbf{\text{P(B)}}& \mathbf{\text{P(A}}\mathbf{\cap }\mathbf{\text{B)}}& \mathbf{\text{P(A}}\mathbf{\cup }\mathbf{\text{B)}}\\ \mathbf{\text{(i)}}& \frac{\mathbf{\text{1}}}{\mathbf{\text{3}}}& \frac{\mathbf{\text{1}}}{\mathbf{\text{5}}}& \frac{\mathbf{\text{1}}}{\mathbf{\text{15}}}& \mathbf{.}\mathbf{\dots \dots }\\ \mathbf{\text{(ii)}}& \mathbf{\text{0.35}}& \mathbf{.}\mathbf{\dots \dots }& \mathbf{\text{0.25}}& \mathbf{\text{0.6}}\\ \mathbf{\text{(iii)}}& \mathbf{\text{0.5}}& \mathbf{\text{0.35}}& \mathbf{.}\mathbf{\dots \dots }& \mathbf{\text{0.7}}\end{array}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}P}\left(\text{A}\cup \text{B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)\\ =\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\\ =\frac{5+3-1}{15}\\ =\frac{7}{15}\\ \left(\mathrm{ii}\right)\text{\hspace{0.33em}P}\left(\text{A}\cup \text{B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)\\ 0.6=0.35+\text{P}\left(\text{B}\right)-0.25\\ 0.6-0.1=\text{P}\left(\text{B}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}\left(\text{B}\right)=0.5\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\text{A}\cup \text{B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)\\ 0.7=0.5+0.35-\text{P}\left(\text{A}\cap \text{B}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}\left(\text{A}\cap \text{B}\right)=0.85-0.7\\ =0.15\end{array}$

Q.31

$\begin{array}{l}\mathbf{\text{Given P(A) =}}\frac{\mathbf{\text{3}}}{\mathbf{\text{5}}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}and P(B) =}}\frac{\mathbf{\text{1}}}{\mathbf{\text{5}}}\mathbf{\text{. Find P(A or B), if A and B are}}\\ \mathbf{\text{mutually exclusive events.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\text{A and B are mutually exclusive events. So,}\\ \text{A}\cap \text{B}=\mathrm{\varphi }\Rightarrow \mathrm{P}(\text{A}\cap \text{B})=0\\ \because \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =\frac{3}{5}+\frac{1}{5}-0\\ =\frac{4}{5}\\ \mathrm{Therefore},\text{P}\left(\mathrm{A}\mathrm{or}\mathrm{B}\right)\text{is}\frac{4}{5}.\end{array}$

Q.32

$\begin{array}{l}\mathbf{\text{If E and F are events such that P(E) =}}\frac{\mathbf{\text{1}}}{\mathbf{\text{4}}}\mathbf{\text{,\hspace{0.17em}P(F) =}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\mathbf{\text{and}}\\ \mathbf{\text{P(E and F) =}}\frac{\mathbf{\text{1}}}{\mathbf{\text{8}}}\mathbf{\text{, find}}\left(\text{i}\right)\mathbf{\text{P}}\left(\text{E or F}\right)\mathbf{\text{,}}\left(\text{ii}\right)\mathbf{\text{P}}\left(\text{not E and not F}\right)\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{P}\left(\mathrm{E}\right)=\frac{1}{4},\mathrm{P}\left(\mathrm{F}\right)=\frac{1}{2}\text{and P}(\mathrm{E}\cap \mathrm{F})=\frac{1}{8}\\ \left(\mathrm{i}\right)\because \mathrm{P}(\mathrm{E}\cup \mathrm{F})=\mathrm{P}\left(\mathrm{E}\right)+\mathrm{P}\left(\mathrm{F}\right)-\mathrm{P}(\mathrm{E}\cap \mathrm{F})\\ =\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\\ =\frac{2+4-1}{8}\\ =\frac{5}{8}\\ \mathrm{Therefore},\text{P}\left(\mathrm{E}\mathrm{or}\mathrm{F}\right)\text{is}\frac{5}{8}.\\ \begin{array}{l}\left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{not}\mathrm{E}\text{and not F}\right)=\mathrm{P}(\mathrm{E}‘\cap \mathrm{F}‘)\\ =\mathrm{P}(\mathrm{E}\cup \mathrm{F})‘\left[\mathrm{By}\text{Demorgan’s Law}\right]\\ =1-\mathrm{P}(\mathrm{E}\cap \mathrm{F})\\ =1-\frac{5}{8}\\ =\frac{8-5}{8}\\ =\frac{3}{8}\\ \mathrm{Therefore},\text{P}\left(\mathrm{not}\mathrm{E}\text{and not F}\right)\mathrm{is}\text{equal to}\frac{3}{8}.\end{array}\end{array}$

Q.33 Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.

Ans

$\begin{array}{l}\because \mathrm{P}\left(\mathrm{not}\mathrm{E}\mathrm{or}\mathrm{not}\mathrm{F}\right)=0.25\\ \Rightarrow \mathrm{P}(\mathrm{E}‘\cup \mathrm{F}‘)=0.25\\ \Rightarrow \mathrm{P}(\mathrm{E}\cap \mathrm{F})‘=0.25\left[\mathrm{By}\text{Demorgarn’s law}\right]\\ \Rightarrow 1-\mathrm{P}(\mathrm{E}\cap \mathrm{F})=0.25[\because \mathrm{P}(\mathrm{E}‘)=1-\mathrm{P}\left(\mathrm{E}\right)]\\ \Rightarrow \mathrm{P}(\mathrm{E}\cap \mathrm{F})=1-0.25\\ \Rightarrow \mathrm{P}(\mathrm{E}\cap \mathrm{F})=0.75\\ \Rightarrow \mathrm{E}\cap \mathrm{F}\ne 0\\ \mathrm{So},\text{\hspace{0.17em}E and F are not mutually exclusive.}\end{array}$

Q.34 A and B are events such that P(A) = 0.42,
P(B) = 0.48 and P(A and B) = 0.16.
Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)

Ans

$\begin{array}{l}\mathrm{Given}:\text{P}\left(\text{A}\right)=0.\text{42},\mathrm{}\text{P}\left(\text{B}\right)=0.\text{48 and P}(\text{A}\cap \text{B})=0.\text{16}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{not}\mathrm{A}\right)=\mathrm{P}(\mathrm{A}‘)\\ =1-\mathrm{P}\left(\mathrm{A}\right)\\ =1-0.42\\ =0.58\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{not}\mathrm{B}\right)=\mathrm{P}(\mathrm{B}‘)\\ =1-\mathrm{P}\left(\mathrm{B}\right)\\ =1-0.\text{48}\\ =0.52\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}\mathrm{or}\mathrm{B}\right)=\mathrm{P}(\mathrm{A}\cup \mathrm{B})\\ =\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =0.\text{42}+0.\text{48}-0.\text{16}\\ =0.74\end{array}$

Q.35 In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Ans

$\begin{array}{l}\text{P}\left(\text{Mathematics}\right)\text{\hspace{0.17em}=\hspace{0.17em}40\%}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.17em}P}\left(\text{M}\right)\text{\hspace{0.17em}=\hspace{0.17em}0.40}\\ \text{P}\left(\text{Biology}\right)\text{\hspace{0.17em}=\hspace{0.17em}30\%}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}P}\left(\text{B}\right)\text{\hspace{0.17em}=\hspace{0.17em}0.30}\end{array}$ $\begin{array}{l}\text{P}\left(\mathrm{Mathematics}\text{and Biology}\right)=10\text{\%}\\ \text{P}\left(\text{M}\cap \text{B}\right)=0.10\\ \text{P}\left(\mathrm{Mathematics}\text{or Biology}\right)=\text{P}\left(\text{M}\cup \text{B}\right)\\ =\text{P}\left(\text{M}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{M}\cap \text{B}\right)\\ =0.40+0.30-0.10\\ =0.60\\ \mathrm{Therefore},\text{probability that student wiall be studying}\\ \text{Mathematics or Biology is 0.60.}\end{array}$

Q.36 In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

Ans

$\begin{array}{l}\mathrm{Let}\text{probability of passing in both exa}\min \text{ations \hspace{0.17em}}\mathrm{be}\mathrm{P}\left(\mathrm{A}\right)\text{and}\\ \text{P}\left(\mathrm{B}\right)\text{respectively.Then, we have}\\ \text{P}\left(\mathrm{A}\right)=0.8,\text{P}\left(\mathrm{B}\right)=0.7\text{and P}(\mathrm{A}\cup \mathrm{B})=0.95\\ \because \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cup \mathrm{B})\\ =0.8+0.7-0.95\\ =0.55\\ \mathrm{Thus},\text{the probability of passing in both the examination is 0.55.}\end{array}$

Q.37 The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

Ans

$\begin{array}{l}\text{Let\hspace{0.17em}the probability that a student will pass the}\mathrm{}\text{final examination}\\ \text{in both English and Hindi}=\mathrm{P}(\mathrm{E}\cap \mathrm{H})\\ \Rightarrow \mathrm{P}(\mathrm{E}\cap \mathrm{H})=0.5\\ \text{Let\hspace{0.17em}the probability that a student will not pass the}\mathrm{}\text{final}\\ \text{examination in neither English nor Hindi}=\mathrm{P}(\mathrm{E}‘\cap \mathrm{H}‘)\\ \Rightarrow \mathrm{P}(\mathrm{E}‘\cap \mathrm{H}‘)=0.1\\ \Rightarrow \mathrm{P}(\mathrm{E}\cup \mathrm{H})‘=0.1\left[\mathrm{By}\text{Demorgan’s law}\right]\\ \Rightarrow 1-\mathrm{P}(\mathrm{E}\cup \mathrm{H})=0.1\\ \Rightarrow \mathrm{P}(\mathrm{E}\cup \mathrm{H})=1-0.1\\ \Rightarrow \mathrm{P}(\mathrm{E}\cup \mathrm{H})=0.9\\ \text{The probability of passing the English}\mathrm{}\text{examination}\\ =\text{P}\left(\mathrm{E}\right)\\ =0.75\\ \because \mathrm{P}(\mathrm{E}\cup \mathrm{H})=\mathrm{P}\left(\mathrm{E}\right)+\mathrm{P}\left(\mathrm{H}\right)-\mathrm{P}(\mathrm{E}\cap \mathrm{H})\\ \Rightarrow 0.9=0.75+\mathrm{P}\left(\mathrm{H}\right)-0.5\\ \Rightarrow \mathrm{P}\left(\mathrm{H}\right)=0.9-0.75+0.5\\ =0.65\\ \text{Therefore, the probability of passing the Hindi}\\ \text{examination is 0.65.}\end{array}$

Q.38 In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.

Ans

$\begin{array}{l}\mathrm{Number}\text{of students in a class}=60\\ \mathrm{Number}\text{of students in NCC}=30\\ \mathrm{Number}\text{of students in NSS}=32\\ \mathrm{Number}\text{of students in NCC and NSS}=24\\ \therefore \mathrm{P}\left(\mathrm{NCC}\right)=\frac{30}{60}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{NSS}\right)=\frac{32}{60}=\frac{8}{15}\\ \mathrm{P}(\mathrm{NCC}\cap \mathrm{NSS})=\frac{24}{60}=\frac{6}{15}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{NCC}\text{or NSS}\right)=\mathrm{P}(\mathrm{NCC}\cup \mathrm{NSS})\\ =\mathrm{P}\left(\mathrm{NCC}\right)+\mathrm{P}\left(\mathrm{NSS}\right)-\mathrm{P}(\mathrm{NCC}\cap \mathrm{NSS})\\ =\frac{1}{2}+\frac{8}{15}-\frac{6}{15}\\ =\frac{15+16-12}{30}\\ =\frac{19}{30}\\ \left(\mathrm{ii}\right)\hspace{0.17em}\mathrm{P}\left(\mathrm{Neither}\text{NCC nor NSS}\right)\\ =\mathrm{P}\{\left(\mathrm{NCC}\right)‘\cup \left(\mathrm{NSS}\right)‘\}\\ =\mathrm{P}(\mathrm{NCC}\cap \mathrm{NSS})‘\\ =1-\mathrm{P}(\mathrm{NCC}\cup \mathrm{NSS})\\ \begin{array}{l}=1-\frac{19}{30}\\ =\frac{11}{30}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{NSS}\text{but not NCC}\right)=\mathrm{P}\{\left(\mathrm{NSS}\right)\cap \left(\mathrm{NCC}\right)‘\}\\ =\mathrm{P}\left(\mathrm{NSS}\right)-\mathrm{P}(\mathrm{NSS}\cap \mathrm{NCC})\\ =\frac{8}{15}-\frac{6}{15}\\ =\frac{2}{15}\end{array}\end{array}$

Q.39 A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

(i) all will be blue?
(ii) at least one will be green?

Ans

$\begin{array}{l}\mathrm{Number}\text{of red balls in box}=10\\ \mathrm{Number}\text{of blue balls in box}=20\\ \mathrm{Number}\text{of green balls in box}=30\\ \mathrm{Total}\text{number of balls in box}=60\\ \mathrm{Number}\text{of ways to select 5 balls \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={}^{60}{\mathrm{C}}_5\\ \left(\mathrm{i}\right)\mathrm{Number}\text{of ways to select 5 blue balls \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={}^{20}{\mathrm{C}}_5\\ \mathrm{P}\left(5\mathrm{blue}\mathrm{balls}\right)=\frac{{}^{20}{\mathrm{C}}_5}{{}^{60}{\mathrm{C}}_5}\\ \left(\mathrm{ii}\right)\text{P}\left(\mathrm{at}\text{least one ball will be green}\right)\\ =1-\mathrm{P}\left(\mathrm{no}\text{green ball}\right)\\ =1-\frac{{}^{30}{\mathrm{C}}_5}{{}^{60}{\mathrm{C}}_5}\end{array}$

Q.40 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Ans

$\begin{array}{l}\mathrm{Total}\text{cards in well-shuffled deck}=\text{52}\\ \text{Number of drawn cards from deck}=4\\ \mathrm{Number}\text{of diamonds in deck}=13\\ \mathrm{Number}\text{of spade in deck}=13\\ \text{P}\left(3\mathrm{diamonds}\text{and 1 spade}\right)=\frac{{}^{13}{\text{C}}_3\times {}^{13}{\text{C}}_1}{{}^{52}{\text{C}}_4}\end{array}$

Q.41 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3)

Ans

$\begin{array}{l}\mathrm{Possible}\text{events in throwing a given dice,S}\left(\mathrm{E}\right)=\{1,1,2,2,2,3\}\\ \left(\mathrm{i}\right)\mathrm{P}\left(2\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{S}\left(\mathrm{E}\right)}\\ =\frac{3}{6}\\ =\frac{1}{2}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(1\mathrm{or}3\right)=\mathrm{P}\left(1\right)+\mathrm{P}\left(3\right)\\ \begin{array}{l}=\frac{2}{6}+\frac{1}{6}\\ =\frac{3}{6}\\ =\frac{1}{2}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{not}3\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{S}\left(\mathrm{E}\right)}\\ =\frac{5}{6}\end{array}\end{array}$

Q.42 In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded.
What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.

Ans

$\begin{array}{l}\mathrm{Total}\text{number of lottery tickets}=10,000\\ \mathrm{Total}\text{number of equal prizes}=10\\ \mathrm{Number}\text{of tickets not having prize}=10,000-10\\ =9990\\ \left(\mathrm{a}\right)\mathrm{P}\left(1\text{prize ticket}\right)=\frac{{}^{10}{\mathrm{C}}_1}{{}^{10000}{\mathrm{C}}_1}\\ =\frac{10}{10,000}\\ =\frac{1}{1000}\\ \mathrm{P}\left(1\text{ticket of without prize}\right)=1-\mathrm{P}\left(1\text{prize ticket}\right)\end{array}$ $\begin{array}{l}=1-\frac{1}{1000}\\ \mathrm{P}\left(1\text{ticket of without prize}\right)=\frac{999}{1000}\\ \left(\mathrm{b}\right)\mathrm{P}\left(2\text{tickets of without prize}\right)=\frac{{}^{9990}{\mathrm{C}}_2}{{}^{10000}{\mathrm{C}}_2}\\ \left(\mathrm{c}\right)\mathrm{P}\left(10\text{tickets of without prize}\right)=\frac{{}^{9990}{\mathrm{C}}_{10}}{{}^{10000}{\mathrm{C}}_{10}}\end{array}$

Q.43 Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that

(a) you both enter the same section?
(b) you both enter the different sections?

Ans

$\begin{array}{l}\mathrm{Number}\text{of students in two sections}=100\\ \mathrm{Number}\text{of students in section A}=40\\ \mathrm{Number}\text{of students in section B}=60\\ \mathrm{Total}\text{number of ways to fill both section by students}\\ ={}^{100}{\mathrm{C}}_{40}\times {}^{60}{\mathrm{C}}_{60}\\ =\frac{100!}{40!\times 60!}\times 1\\ =\frac{100!}{40!\times 60!}\\ \left(\mathrm{a}\right)\text{When both enter in the same section:}\\ \text{Probability}\left(\mathrm{Both}\text{are in same section}\right)=\frac{{}^{98}{\mathrm{C}}_{38}\times {}^{60}{\mathrm{C}}_{60}+{}^{98}{\mathrm{C}}_{58}\times {}^{40}{\mathrm{C}}_{40}}{\left(\frac{100!}{40!\times 60!}\right)}\end{array}$ $\begin{array}{l}=\frac{\frac{98!}{38!60!}\times 1+\frac{98!}{58!40!}\times 1}{\left(\frac{100!}{40!\times 60!}\right)}\\ =\frac{\frac{98!}{38!58!}(\frac{1}{60\times 59}+\frac{1}{40\times 39})}{\left(\frac{100\times 99\times 98!}{40\times 39\times 38!\times 60\times 59\times 58!}\right)}\\ =\frac{\left(\frac{40\times 39+60\times 59}{60\times 59\times 40\times 39}\right)}{\left(\frac{100\times 99}{40\times 39\times 60\times 59}\right)}\\ =\frac{1560+3540}{9900}\\ =\frac{5100}{9900}\\ =\frac{51}{99}\\ =\frac{17}{33}\\ \mathrm{Therefore},\text{when both students are in the same section.}\\ \text{Then,}\mathrm{the}\text{required probability is}\frac{17}{33}.\\ \left(\mathrm{b}\right)\mathrm{P}\left(\mathrm{When}\text{both are not in same section}\right)\\ =1-\text{Probability}\left(\begin{array}{l}\mathrm{Both}\text{are in}\\ \text{same section}\end{array}\right)\\ =1-\frac{17}{33}=\frac{16}{33}\end{array}$ $\text{Then,}\mathrm{the}\text{required probability is}\frac{16}{33}.$

Q.44 Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Ans

Number of letters = 3
Number of persons = 3
Total number of ways to insert 3
letters in 3 envelops = 3!
= 6

Number of ways of inserting 3 letters in 3 envelops
so that no letter is in proper envelope = 1
Total number of ways of inserting 3 letters in 3
envelopes = 3! – one letter is in proper
envelopes and two letters are in wrong envelops – one letter in right envelopes and two letters are in
wrong envelopes – Number of ways in which all are in proper envelopes
= 6 – ³C₁ x 1 – 1
= 6 – 3 – 1
= 2

$\begin{array}{l}\therefore \mathrm{P}\left(\mathrm{at}\text{least one letter is in its proper envelope}\right)\\ \mathrm{}=1-\mathrm{P}\left(\mathrm{No}\text{letter is in its proper envelope}\right)\\ \mathrm{}=1-\frac{2}{6}\end{array}$ $\begin{array}{l}\mathrm{}=1-\frac{1}{3}\\ \mathrm{}=\frac{2}{3}\\ \mathrm{Thus},\text{the required propbability is}\frac{2}{3}.\end{array}$

Q.45 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
Find (i) P (A∪B) (ii) P(A´∩B´) (iii) P(A ∩ B´) (iv) P(B ∩ A´)

Ans

$\begin{array}{l}\mathrm{We}\text{have, P}\left(\mathrm{A}\right)=0.\text{54},\mathrm{}\text{\hspace{0.17em}P}\left(\text{B}\right)=0.\text{69 and P}(\text{A}\cap \text{B})=0.\text{35}\\ \left(\mathrm{i}\right)\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =0.54+0.69-0.35\\ =0.88\\ \left(\mathrm{ii}\right)\mathrm{P}(\mathrm{A}‘\cap \mathrm{B}‘)=\mathrm{P}(\mathrm{A}\cup \mathrm{B})‘\left[\mathrm{By}\text{Demorgan’s theorem}\right]\\ =1-\mathrm{P}(\mathrm{A}\cup \mathrm{B})\\ =1-0.88\\ =0.12\\ \left(\mathrm{iii}\right)\mathrm{P}(\mathrm{A}\cap \mathrm{B}‘)=\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =0.54-0.35\\ =0.19\\ \left(\mathrm{iv}\right)\mathrm{P}(\mathrm{B}\cap \mathrm{A}‘)=\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =0.69-0.35\\ =0.34\end{array}$

Q.46 From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

Sr. No.	Name	Sex	Age in years
1	Harish	M	30
2	Rohan	M	46
3	Sheetal	F	46
4	Alis	F	28
5	Salim	M	41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

Ans

$\begin{array}{l}\mathrm{Number}\text{of male}=\text{3}\\ \text{Number of female over 35 years}=\text{1}\\ \text{Total number of people}=\text{5}\\ \text{P}\left(\mathrm{Male}\text{or over 35 years people}\right)=\frac{4}{5}\end{array}$

Q.47 If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated?
(ii) the repetition of digits is not allowed?

Ans

$\begin{array}{l}\mathrm{Given}\text{digits are 0,1,3,5 and 7.}\\ \left(\mathrm{i}\right)\mathrm{When}\text{digits are repeated.}\\ \text{Total number of 4 digits number greater}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or equal to 5000}=2\times 5\times 5\times 5\\ =250\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{number of 4-digits number greater than 5000}\\ =250-1\\ =249\\ \mathrm{A}\text{number is divisible by 5 if its one’s digit is either 0 or 5.}\\ \text{Therefore, 4-digits number formed by given digits and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}greater or equal to 5000}=2\times 5\times 5\times 2\\ =100\\ \mathrm{Therefore},\text{number of 4-digits number greater than 5000}\\ =100-1\\ =99\\ \mathrm{P}\left(\mathrm{When}\text{digits are repeated}\right)=\frac{99}{249}\\ =\frac{33}{83}\\ \left(\mathrm{ii}\right)\mathrm{When}\text{repeatition of digits is not allowed.}\\ \text{Total number of 4-digit numbers formed from the digits}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0,1,3,5 and 7}=2\times 4\times 3\times 2\\ =48\\ \mathrm{Number}\text{of 4-digit number divisible by 5:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}When one’s digit is 0}=2\times 3\times 2\times 1\\ =12\\ \mathrm{Number}\text{of 4-digit number divisible by 5:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}When one’s digit is 5}=1\times 3\times 2\times 1\\ =6\\ \mathrm{Total}\text{numbers divisible by 5}=12+6\\ =18\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{P}\left(\mathrm{When}\text{digits are not repeated}\right)\\ =\frac{18}{48}\\ =\frac{3}{8}\\ \mathrm{Hence},\text{probability of forming a number divisible by 5 is}\frac{3}{8}.\end{array}$

Q.48 The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Ans

$\begin{array}{l}\mathrm{Number}\text{of ways to select 4 digits out of 10 digits}={}^{10}{\mathrm{C}}_4\\ =\frac{10!}{4!6!}\\ =\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\\ =210\\ \mathrm{Number}\text{of ways to arrange 4 selected digits}=210\times 4!\\ =5040\\ \mathrm{Since},\text{there is only one sequence to open the lock.}\\ \text{\hspace{0.17em}P}\left(\mathrm{right}\text{sequence to open the lock}\right)=\frac{1}{5040}\end{array}$