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

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