# NCERT Solutions For Class 12 Maths Chapter 13 Probability (Ex 13.4) Exercise 13.4

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## NCERT Solutions For Class 12 Maths Chapter 13 Probability (Ex 13.4) Exercise 13.4

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### Access NCERT Solutions For Class 12 Maths Chapter 13 – Probability

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### NCERT Solution Class 12 Maths Of Chapter 13 All Exercises

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## NCERT Solutions For Class 12 Maths Chapter 13 Probability Exercise 13.4

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### Important Topics Covered In Exercise 13.4 Of NCERT Solutions For Class 12 Maths Chapter 13 Probability

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### Below Are Some Key Points We Will Learn From This Chapter

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Learning should never be stopped, hence students should learn from each chapter and note it down to ensure that their learning will last a lifetime.Mentioned below are some important points that students will learn from the chapter.

1. The meaning of Probability.
2. Different types of Probability.
3. Examples of Probability.
4. Features of Probability.

Q.1 State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i)

 X 0 1 2 P(X) 0.4 0.4 0.2

(ii)

 X 0 1 2 3 4 P(X) 0.1 0.5 0.2 – 0.1 0.3

(iii)

 Y –1 0 1 P(Y) 0.6 0.1 0.2

(iv)

 Z 3 2 1 0 –1 P(Z) 0.3 0.2 0.4 0.1 0.05

Ans
It is known that the sum of all the probabilities in a probability distribution is one.
(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1
Therefore, the given table is a probability distribution of random variables.
(ii) It can be seen that for X = 3, P (X) = −0.1
It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.
(iii) Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Therefore, the given table is not a probability distribution of random variables.
(iv) Sum of the probabilities
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
= 1.05 ≠ 1
Therefore, the given table is not a probability distribution of random variables.

Q.2 An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

Ans
The two balls selected can be represented as BB, BR, RB, RR where B represents a black ball and R represents a red ball. X represents the number of black balls.
So, X (BB) = 2
X(BR) = 1
X(RB) = 1
X(RR) = 0
Therefore, the possible values of X are 0, 1, and 2.
Yes, X is a random variable.

Q.3 Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Ans

$\begin{array}{l}\mathrm{A}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{six}\mathrm{times}\mathrm{and}\mathrm{X}\mathrm{represents}\mathrm{the}\mathrm{difference}\\ \mathrm{between}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}\mathrm{and}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{tails}.\\ \therefore \mathrm{X}\left[6\mathrm{H},0\mathrm{T}\right]=\mathrm{X}\left[6\mathrm{times}\mathrm{H}\mathrm{and}0\mathrm{times}\mathrm{T}\right]=|6-0|=1\\ \mathrm{X}\left[5\mathrm{H},1\mathrm{T}\right]=\mathrm{X}\left[5\mathrm{times}\mathrm{H}\mathrm{and}1\mathrm{times}\mathrm{T}\right]=|5-1|=4\\ \mathrm{X}\left[4\mathrm{H},2\mathrm{T}\right]=\mathrm{X}\left[4\mathrm{times}\mathrm{H}\mathrm{and}2\mathrm{times}\mathrm{T}\right]=|4-2|=2\\ \mathrm{X}\left[3\mathrm{H},3\mathrm{T}\right]=\mathrm{X}\left[3\mathrm{times}\mathrm{H}\mathrm{and}3\mathrm{times}\mathrm{T}\right]=|3-3|=0\\ \mathrm{X}\left[2\mathrm{H},4\mathrm{T}\right]=\mathrm{X}\left[2\mathrm{times}\mathrm{H}\mathrm{and}4\mathrm{times}\mathrm{T}\right]=|2-4|=2\\ \mathrm{X}\left[1\mathrm{H},5\mathrm{T}\right]=\mathrm{X}\left[1\mathrm{times}\mathrm{H}\mathrm{and}5\mathrm{times}\mathrm{T}\right]=|1-5|=4\\ \mathrm{X}\left[0\mathrm{H},6\mathrm{T}\right]=\mathrm{X}\left[0\mathrm{times}\mathrm{H}\mathrm{and}6\mathrm{times}\mathrm{T}\right]=|0-6|=6\\ \mathrm{Therefore},\mathrm{the}\mathrm{possible}\mathrm{values}\mathrm{of}\mathrm{X}\mathrm{are}0, 1, 2, 4\mathrm{and}6\mathrm{.}\end{array}$

Q.4 Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{When}\mathrm{one}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{twice},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \left\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\right\}\\ \mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{heads}.\\ \therefore \mathrm{X}\left(\mathrm{HH}\right)=2,\mathrm{ }\mathrm{X}\left(\mathrm{HT}\right)=1,\mathrm{}\mathrm{X}\left(\mathrm{TH}\right)=1,\mathrm{}\mathrm{X}\left(\mathrm{TT}\right)=0\\ \mathrm{Therefore},\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1,\mathrm{or}2\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{HH}\right)=\frac{1}{4},\mathrm{ }\mathrm{P}\left(\mathrm{TH}\right)=\frac{1}{4},\mathrm{ }\mathrm{P}\left(\mathrm{HT}\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(\mathrm{TT}\right)=\frac{1}{4}\\ \mathrm{Then},\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{TT}\right)=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HT}\right)+\mathrm{P}\left(\mathrm{TH}\right)\\ =\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HH}\right)\\ =\frac{1}{4}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 P(X) $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{4}$

$\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{When}\mathrm{three}\mathrm{coins}\mathrm{are}\mathrm{tossed}\mathrm{simultaneously},\mathrm{the}\mathrm{sample}\mathrm{space}\\ \mathrm{is} \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THT},\mathrm{THH},\mathrm{TTH},\mathrm{TTT}\right\}\\ \mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{tails}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1, 2,\mathrm{or}3\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{HHH}\right)=\frac{1}{8}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)\\ =\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HTT}\right)+\mathrm{P}\left(\mathrm{TTH}\right)+\mathrm{P}\left(\mathrm{THT}\right)\\ =\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{TTT}\right)\\ =\frac{1}{8}\\ \mathrm{Thus},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 3 P(X) $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$

$\begin{array}{l}\left(\mathrm{iii}\right)\mathrm{When}\mathrm{a}\mathrm{coin}\mathrm{is}\mathrm{tossed}\mathrm{four}\mathrm{times},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \mathrm{S}=\left\{\begin{array}{l}\mathrm{HHHH},\mathrm{HHHT},\mathrm{HHTH},\mathrm{HTHH},\mathrm{THHH},\mathrm{HHTT},\mathrm{HTTH},\mathrm{TTHH},\\ \mathrm{HTHT}, \mathrm{THTH},\mathrm{THHT}, \mathrm{HTTT}, \mathrm{THTT},\mathrm{TTHT},\mathrm{TTTH}, \mathrm{TTTT}\end{array}\right\}\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable},\mathrm{which}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{heads}.\mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1, 2, 3,\mathrm{or}4\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{HHHH}\right)=\frac{1}{16}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HHHT}\right)+\mathrm{P}\left(\mathrm{HHTH}\right)+\mathrm{P}\left(\mathrm{HTHH}\right)+\mathrm{P}\left(\mathrm{THHH}\right)\\ =\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\\ =\frac{4}{16}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HHTT}\right)+\mathrm{P}\left(\mathrm{HTTH}\right)+\mathrm{P}\left(\mathrm{TTHH}\right)+\mathrm{P}\left(\mathrm{HTHT}\right)+\mathrm{P}\left(\mathrm{THTH}\right)\\ +\mathrm{ }\mathrm{P}\left(\mathrm{THHT}\right)\\ =\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\\ =\frac{6}{16}=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{HTTT}\right)+\mathrm{P}\left(\mathrm{THTT}\right)+\mathrm{P}\left(\mathrm{TTHT}\right)+\mathrm{P}\left(\mathrm{TTTH}\right)\\ =\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}\\ =\frac{4}{16}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(\mathrm{TTTT}\right)=\frac{1}{16}\end{array}$

 X 0 1 2 3 4 P(X) $\frac{1}{16}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{4}$ $\frac{1}{16}$

Q.5 Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die

Ans

$\begin{array}{l}\mathrm{When}\mathrm{a}\mathrm{die}\mathrm{is}\mathrm{tossed}\mathrm{two}\mathrm{times},\mathrm{we}\mathrm{obtain}\left(6×6\right) = 36\mathrm{number}\\ \mathrm{of}\mathrm{observations}.\mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable},\mathrm{which}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{successes}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{Here},\mathrm{success}\mathrm{refers}\mathrm{to}\mathrm{the}\mathrm{number}\mathrm{greater}\mathrm{than}4\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{number}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}4\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{tosses}\right)\\ \mathrm{ }=\frac{4}{6}×\frac{4}{6}=\frac{4}{9}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{number}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}4\mathrm{on}\mathrm{first}\mathrm{toss}\mathrm{and}\\ \mathrm{greater}\mathrm{than}4\mathrm{on}\mathrm{second}\mathrm{toss}\right) +\mathrm{P}\left(\mathrm{number}\mathrm{greater}\\ \mathrm{than}4\mathrm{on}\mathrm{first}\mathrm{toss}\mathrm{and}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}4\mathrm{on}\\ \mathrm{second}\mathrm{toss}\right)\\ \mathrm{ }=\frac{4}{6}×\frac{2}{6}+\frac{2}{6}×\frac{4}{6}=\frac{4}{9}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{number}\mathrm{greater}\mathrm{than}4\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{tosses}\right)\\ \mathrm{ }=\frac{2}{6}×\frac{2}{6}=\frac{1}{9}\\ \mathrm{Thus},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 P(X) $\frac{4}{9}$ $\frac{4}{9}$ $\frac{1}{9}$

$\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{Here},\mathrm{success}\mathrm{means}\mathrm{six}\mathrm{appears}\mathrm{on}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{die}\mathrm{.}\\ \mathrm{P}\mathrm{}\left(\mathrm{Y}=0\right)=\mathrm{P}\left(\mathrm{six}\mathrm{}\mathrm{does}\mathrm{}\mathrm{not}\mathrm{}\mathrm{appear}\mathrm{}\mathrm{on}\mathrm{}\mathrm{any}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{dice}\right)\\ =\frac{5}{6}×\frac{5}{6}=\frac{25}{36}\\ \mathrm{P}\left(\mathrm{Y}=1\right)=\mathrm{P}\mathrm{ }\left(\mathrm{six}\mathrm{}\mathrm{appears}\mathrm{}\mathrm{on}\mathrm{}\mathrm{at}\mathrm{}\mathrm{least}\mathrm{}\mathrm{one}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{dice}\right)=\frac{11}{36}\\ \mathrm{Thus},\mathrm{}\mathrm{the}\mathrm{}\mathrm{required}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{distribution}\mathrm{}\mathrm{is}\mathrm{}\mathrm{as}\mathrm{}\mathrm{follows}:\end{array}$

 Y 0 1 P(Y) $\frac{25}{36}$ $\frac{11}{36}$

Q.6 From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{out}\mathrm{of}30\mathrm{bulbs}, 6\mathrm{are}\mathrm{defective}\mathrm{.}\\ ⇒\mathrm{Number}\mathrm{of}\mathrm{non}–\mathrm{defective}\mathrm{bulbs}=\mathrm{30}-\mathrm{6}=\mathrm{24}\\ \mathrm{P}\left(\mathrm{Non}-\mathrm{defective}\mathrm{bulb}\right)=\frac{24}{30}=\frac{6}{5}\\ \mathrm{P}\left(\mathrm{Defective}\mathrm{bulb}\right)=\frac{6}{30}=\frac{1}{5}\\ 4\mathrm{bulbs}\mathrm{are}\mathrm{drawn}\mathrm{from}\mathrm{the}\mathrm{lot}\mathrm{with}\mathrm{replacement}\mathrm{.}\\ \mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{that}\mathrm{denotes}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{defective}\mathrm{bulbs}\mathrm{in}\mathrm{the}\mathrm{selected}\mathrm{bulbs}\mathrm{.}\\ \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(4\mathrm{non}–\mathrm{defective}\mathrm{and}0\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{0}^{4}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(3\mathrm{non}–\mathrm{defective}\mathrm{and}1\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{1}^{4}.\left(\frac{1}{5}\right).{\left(\frac{4}{5}\right)}^{3}=\frac{256}{625}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(2\mathrm{non}–\mathrm{defective}\mathrm{and}2\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{2}^{4}.{\left(\frac{1}{5}\right)}^{2}.{\left(\frac{4}{5}\right)}^{2}=\frac{96}{625}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(1\mathrm{non}–\mathrm{defective}\mathrm{and}3\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{3}^{4}.{\left(\frac{1}{5}\right)}^{3}.\left(\frac{4}{5}\right)=\frac{16}{625}\\ \mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(0\mathrm{non}–\mathrm{defective}\mathrm{and}4\mathrm{defective}\right)\\ \mathrm{ }=\mathrm{C}_{4}^{4}.{\left(\frac{1}{5}\right)}^{4}.{\left(\frac{4}{5}\right)}^{0}=\frac{1}{625}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 3 4 P(X) $\frac{256}{625}$ $\frac{256}{625}$ $\frac{96}{625}$ $\frac{16}{625}$ $\frac{1}{625}$

Q.7 A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Ans

$\begin{array}{l}\text{Let the probability of getting a tail in the biased coin be x}\text{.}\\ \therefore \text{}\text{\hspace{0.17em}}\text{}\text{}P\left(T\right)=x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(H\right)=3x\\ Since,\text{P}\left(T\right)+P\left(H\right)=1\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+3x=1\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x=1\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{1}{4}\\ Therefore,\text{\hspace{0.17em}}P\left(T\right)=\frac{1}{4}\text{and P}\left(H\right)=\frac{3}{4}\\ \text{When the coin is tossed twice, the sample space is}\\ \text{{HH, TT, HT, TH}}\text{.}\\ \text{Let X be the random variable representing the number of tails}\text{.}\\ \therefore P\left(X=0\right)=P\left(no\text{tail}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=P\left(H\right)×P\left(H\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}×\frac{3}{4}=\frac{9}{16}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(X=1\right)=P\left(one\text{tail}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=P\left(HT\right)+P\left(TH\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}×\frac{1}{4}+\frac{1}{4}×\frac{3}{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{6}{16}=\frac{3}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(X=2\right)=P\left(two\text{tail}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=P\left(TT\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}×\frac{1}{4}=\frac{1}{16}\\ \end{array}$

Therefore, the required probability distribution is as follows:

 X 0 1 2 P(X) $\frac{9}{16}$ $\frac{3}{8}$ $\frac{1}{16}$

Q.8 A random variable X has the following probability distribution:

 X 0 1 2 3 4 5 6 7 P(X) 0 k 2 k 2 k 3 k K2 2k2 7k2+ k

Determine
(i) K (ii) P(X < 3) (iii) P(X > 6) (iv) P(0 < X < 3)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Since}\mathrm{sum}\mathrm{of}\mathrm{probabilities}\mathrm{of}\mathrm{a}\mathrm{probability}\mathrm{distribution}\mathrm{of}\\ \mathrm{random}\mathrm{variables}\mathrm{is}\mathrm{one}\mathrm{.}\\ \mathrm{ }0+\mathrm{k}+2\mathrm{k}+2\mathrm{k}+3\mathrm{ }\mathrm{k}+{\mathrm{k}}^{2}+2\mathrm{ }{\mathrm{k}}^{2}+7\mathrm{ }{\mathrm{k}}^{2}+\mathrm{k}=1\\ ⇒ \mathrm{ }10\mathrm{ }{\mathrm{k}}^{2}+9\mathrm{ }\mathrm{k}-1=0\\ ⇒ \mathrm{ }\left(10\mathrm{ }\mathrm{k}-1\right)\left(\mathrm{k}+1\right)=0\\ ⇒ \mathrm{ }\mathrm{k}=-1, \frac{1}{10}\\ \mathrm{k}=-1\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{as}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{an}\mathrm{event}\mathrm{is}\mathrm{never}\\ \mathrm{negative}\mathrm{.}\\ \therefore \mathrm{ }\mathrm{k}=\frac{1}{10}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{x}<3\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =0+\mathrm{k}+2\mathrm{k}\\ =3\mathrm{k}\\ =3×\frac{1}{10}=\frac{3}{10}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{x}<6\right)=\mathrm{P}\left(\mathrm{X}=7\right)\\ =7\mathrm{ }{\mathrm{k}}^{2}+\mathrm{k}\\ =7{\left(\frac{1}{10}\right)}^{2}+\frac{1}{10}\\ =\frac{7}{100}+\frac{1}{10}\\ =\frac{7+10}{100}=\frac{17}{100}\\ \left(\mathrm{iv}\right)\mathrm{P}\left(0<\mathrm{x}<3\right)\\ =\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{k}+2\mathrm{k}\\ =3\mathrm{k}\\ =3\left(\frac{1}{10}\right)=\frac{3}{10}\end{array}$

Q.9

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{random}\mathrm{}\mathrm{variable}\mathrm{}\mathrm{X}\mathrm{}\mathrm{has}\mathrm{}\mathrm{a}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{distribution}\mathrm{}\mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{following}\mathrm{}\mathrm{form},\mathrm{where}\mathrm{}\mathrm{k}\mathrm{}\mathrm{is}\mathrm{}\mathrm{some}\mathrm{}\mathrm{number}:\\ \mathrm{P}\left(\mathbf{X}\right)=\left\{\begin{array}{l}\mathrm{k},\mathrm{ifx}=0\\ 2\mathrm{k},\mathrm{ifx}=1\\ 3\mathrm{k},\mathrm{ifx}=2\\ 0,\mathrm{otherwise}\end{array}\\ \left(\mathbf{a}\right)\mathrm{Determine}\mathrm{}\mathrm{the}\mathrm{}\mathrm{value}\mathrm{}\mathrm{of}\mathrm{}\mathrm{k}.\\ \left(\mathrm{b}\right)\mathrm{Find}\mathrm{}\mathrm{P}\left(\mathrm{X}<2\right),\mathrm{P}\left(\mathrm{X}\le 2\right),\mathrm{P}\left(\mathrm{X}\ge 2\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Since},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{probabilities}\mathrm{of}\mathrm{a}\mathrm{probability}\mathrm{distribution}\\ \mathrm{of}\mathrm{random}\mathrm{variables}\mathrm{is}\mathrm{one}\mathrm{.}\\ \mathrm{i}.\mathrm{e}.,\mathrm{k}+2\mathrm{k}+3\mathrm{k}+0=1\\ ⇒ 6\mathrm{k}=1\\ ⇒ \mathrm{k}=\frac{1}{6}\\ \left(\mathrm{b}\right)\mathrm{P}\left(\mathrm{x}<2\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{k}+2\mathrm{k}\\ =3\mathrm{k}\\ =3×\frac{1}{6}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{x}\le 2\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{k}+2\mathrm{ }\mathrm{k}+3\mathrm{ }\mathrm{k}\\ =6\mathrm{ }\mathrm{k}\\ =6×\frac{1}{6}=1\\ \mathrm{P}\left(\mathrm{x}\ge 2\right)=\mathrm{P}\left(\mathrm{X}=2\right)+\mathrm{P}\left(\mathrm{X}\ge 2\right)\\ =3\mathrm{ }\mathrm{k}+0\\ =3\mathrm{ }\mathrm{k}\\ =3×\frac{1}{6}\\ =\frac{1}{2}\end{array}$

Q.10 Find the mean number of heads in three tosses of a fair coin.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{denotes}\mathrm{the}\mathrm{success}\mathrm{of}\mathrm{getting}\mathrm{heads}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{sample}\mathrm{space}\mathrm{is}\\ \mathrm{S}= \left\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\right\}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{seen}\mathrm{that}\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1, 2,\mathrm{or}3\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{TTT}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{T}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{8}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{T}\right)+\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{H}\right)+\mathrm{P}\left(\mathrm{T}\right).\mathrm{P}\left(\mathrm{H}\right).\mathrm{P}\left(\mathrm{H}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ \mathrm{ }=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\\ \mathrm{ }=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{HHT}\right)+\mathrm{P}\left(\mathrm{HTH}\right)+\mathrm{P}\left(\mathrm{THH}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ \mathrm{ }=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\\ \mathrm{ }=\frac{3}{8}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(\mathrm{HHH}\right)\\ \mathrm{ }=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ \mathrm{ }=\frac{1}{8}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 3 P(X) null $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$

$\begin{array}{l}\mathrm{Mean}\mathrm{of}\mathrm{X}=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=0×\frac{1}{8}+1×\frac{3}{8}+2×\frac{3}{8}+3×\frac{1}{8}\\ \mathrm{ }=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\\ \mathrm{ }=\frac{12}{8}=1.5\end{array}$

Q.11 Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represents}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{sixes}\mathrm{obtained}\mathrm{when}\mathrm{two}\mathrm{dice}\\ \mathrm{are}\mathrm{thrown}\mathrm{simultaneously}.\\ \mathrm{P}\left(\mathrm{getting}6\right)=\frac{1}{6}\\ \mathrm{P}\left(\mathrm{not}\mathrm{getting}6\right)=\frac{5}{6}\\ \mathrm{Therefore},\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}0, 1,\mathrm{or}2\mathrm{.}\\ \therefore \mathrm{P}\mathrm{}\left(\mathrm{X}=0\right)=\mathrm{P}\left(\mathrm{not}\mathrm{getting}\mathrm{six}\mathrm{on}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{dice}\right)\\ =\frac{5}{6}×\frac{5}{6}=\frac{25}{36}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{first}\mathrm{}\mathrm{die}\mathrm{}\mathrm{and}\mathrm{}\mathrm{no}\mathrm{}\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{second}\mathrm{}\mathrm{die}\right)+\\ \mathrm{ }\mathrm{P}\mathrm{}\left(\mathrm{no}\mathrm{}\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{first}\mathrm{}\mathrm{die}\mathrm{}\mathrm{and}\mathrm{}\mathrm{six}\mathrm{ }\mathrm{on}\mathrm{second}\mathrm{die}\right)\\ \mathrm{ }=\frac{1}{6}×\frac{5}{6}+\frac{5}{6}×\frac{1}{6}=\frac{10}{36}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(\mathrm{six}\mathrm{}\mathrm{on}\mathrm{}\mathrm{both}\mathrm{}\mathrm{dice}\right)\\ \mathrm{ }=\frac{1}{36}\\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{distribution}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 0 1 2 P(X) $\frac{25}{36}$ $\frac{10}{36}$ $\frac{1}{36}$

$\begin{array}{l}\mathrm{Then},\mathrm{expectation}\mathrm{of}\mathrm{X}\\ \mathrm{ }=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=0×\frac{25}{36}+1×\frac{10}{36}+2×\frac{1}{36}\\ \mathrm{ }=\frac{12}{36}=\frac{1}{3}\end{array}$

Q.12 Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).

Ans

$\begin{array}{l}\mathrm{The}\mathrm{two}\mathrm{positive}\mathrm{integers}\mathrm{can}\mathrm{be}\mathrm{selected}\mathrm{from}\mathrm{the}\mathrm{first}\mathrm{six}\\ \mathrm{positive}\mathrm{integers}\mathrm{without}\mathrm{replacement}\mathrm{in}6×\mathrm{5}=30\mathrm{ways}\\ \mathrm{X}\mathrm{represents}\mathrm{the}\mathrm{larger}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{numbers}\mathrm{obtained}\mathrm{.}\\ \mathrm{Therefore},\mathrm{X}\mathrm{can}\mathrm{take}\mathrm{the}\mathrm{value}\mathrm{of}2, 3, 4, 5,\mathrm{or}6\mathrm{.}\\ \mathrm{For}\mathrm{X}=2,\mathrm{the}\mathrm{possible}\mathrm{observations}\mathrm{are}\left(1, 2\right)\mathrm{and}\left(2, 1\right)\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=2\right)=\frac{2}{30}=\frac{1}{15}\\ \mathrm{For}\mathrm{X}=3,\mathrm{the}\mathrm{possible}\mathrm{observations}\mathrm{are}\left(1,3\right), \left(2,3\right), \left(3,1\right)\\ \mathrm{and}\left(3,2\right)\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=3\right)=\frac{4}{30}=\frac{2}{15}\\ \mathrm{For}\mathrm{X}=4,\mathrm{the}\mathrm{possible}\mathrm{observations}\mathrm{are}\left(1, 4\right), \left(2, 4\right), \left(3, 4\right),\\ \left(4, 3\right), \left(4, 2\right)\mathrm{and}\left(4, 1\right)\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=4\right)=\frac{6}{30}=\frac{1}{5}\\ \mathrm{For}\mathrm{}\mathrm{X}=5, \mathrm{the}\mathrm{}\mathrm{possible}\mathrm{}\mathrm{observations}\mathrm{}\mathrm{are}\mathrm{}\left(1,\mathrm{}5\right),\mathrm{}\left(2,\mathrm{}5\right),\mathrm{}\left(3,\mathrm{}5\right),\mathrm{}\\ \left(4,\mathrm{}5\right),\mathrm{}\left(5,\mathrm{}4\right),\mathrm{}\left(5,\mathrm{}3\right),\mathrm{}\left(5,2\right)\mathrm{}\mathrm{and}\mathrm{}\left(5,\mathrm{}1\right).\\ \therefore \mathrm{P}\left(\mathrm{X}=5\right)=\frac{8}{30}=\frac{4}{15}\\ \mathrm{For}\mathrm{}\mathrm{X}=6,\mathrm{the}\mathrm{}\mathrm{possible}\mathrm{}\mathrm{observations}\mathrm{}\mathrm{are}\mathrm{}\left(1,\mathrm{}6\right),\mathrm{}\left(2,\mathrm{}6\right),\mathrm{}\left(3,\mathrm{}6\right),\mathrm{}\\ \left(4,\mathrm{}6\right),\mathrm{}\left(5,\mathrm{}6\right), \left(6,5\right),\left(6,\mathrm{}4\right),\mathrm{}\left(6,3\right),\mathrm{}\left(6,\mathrm{}2\right)\mathrm{}\mathrm{and}\mathrm{}\left(6,\mathrm{}1\right).\\ \therefore \mathrm{P}\left(\mathrm{X}=6\right)=\frac{10}{30}=\frac{1}{3}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 2 3 4 5 6 P(X) $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{4}{15}$ $\frac{1}{3}$

$\begin{array}{l}\mathrm{Then},\mathrm{expectation}\mathrm{of}\mathrm{X}\\ \mathrm{ }=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=2×\frac{1}{15}+3×\frac{2}{15}+4×\frac{1}{5}+5×\frac{4}{15}+6×\frac{1}{3}\\ \mathrm{ }=\frac{2}{15}+\frac{6}{15}+\frac{4}{5}+\frac{20}{15}+\frac{6}{3}\\ \mathrm{ }=\frac{2+6+12+20+30}{15}\\ \mathrm{ }=\frac{70}{15}=\frac{14}{3}\end{array}$

Q.13 Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Ans

$\begin{array}{l}\mathrm{When}\mathrm{two}\mathrm{fair}\mathrm{dice}\mathrm{are}\mathrm{rolled},\\ \mathrm{Number}\mathrm{of}\mathrm{observations}=\mathrm{6}×\mathrm{6}=\mathrm{36}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(1,1\right)\\ =\frac{1}{36}\\ \mathrm{P}\left(\mathrm{X}=3\right)=\mathrm{P}\left(1,2\right)+\mathrm{P}\left(2,1\right)\\ =\frac{2}{36}\\ \mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(2,2\right)+\mathrm{P}\left(3,1\right)+\mathrm{P}\left(1,3\right)\\ =\frac{3}{36}\\ \mathrm{P}\left(\mathrm{X}=5\right)=\mathrm{P}\left(2,3\right)+\mathrm{P}\left(3,2\right)+\mathrm{P}\left(4,1\right)+\mathrm{P}\left(1,4\right)\\ =\frac{4}{36}\\ \mathrm{P}\left(\mathrm{X}=6\right)=\mathrm{P}\left(2,4\right)+\mathrm{P}\left(3,3\right)+\mathrm{P}\left(4,2\right)+\mathrm{P}\left(5,1\right)+\mathrm{P}\left(1,5\right)\\ =\frac{5}{36}\\ \mathrm{P}\left(\mathrm{X}=7\right)=\mathrm{P}\left(2,5\right)+\mathrm{P}\left(3,4\right)+\mathrm{P}\left(4,3\right)+\mathrm{P}\left(5,2\right)+\mathrm{P}\left(1,6\right)+\mathrm{P}\left(6,1\right)\\ =\frac{6}{36}\\ \mathrm{P}\left(\mathrm{X}=8\right)=\mathrm{P}\left(2,6\right)+\mathrm{P}\left(3,5\right)+\mathrm{P}\left(4,4\right)+\mathrm{P}\left(5,3\right)+\mathrm{P}\left(6,2\right)\\ =\frac{5}{36}\\ \mathrm{P}\left(\mathrm{X}=9\right)=\mathrm{P}\left(3,6\right)+\mathrm{P}\left(4,5\right)+\mathrm{P}\left(5,4\right)+\mathrm{P}\left(6,3\right)\\ =\frac{4}{36}\\ \mathrm{P}\left(\mathrm{X}=10\right)=\mathrm{P}\left(4,6\right)+\mathrm{P}\left(5,5\right)+\mathrm{P}\left(6,4\right)\\ =\frac{3}{36}\\ \mathrm{P}\left(\mathrm{X}=11\right)=\mathrm{P}\left(5,6\right)+\mathrm{P}\left(6,5\right)\\ =\frac{2}{36}\\ \mathrm{P}\left(\mathrm{X}=12\right)=\mathrm{P}\left(6,6\right)\\ =\frac{1}{36}\\ \mathrm{Thus},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 2 3 4 5 6 7 8 9 10 11 12 P(X) $\frac{1}{36}$ $\frac{2}{36}$ $\frac{3}{36}$ $\frac{4}{36}$ $\frac{5}{36}$ $\frac{6}{36}$ $\frac{5}{36}$ $\frac{4}{36}$ $\frac{3}{36}$ $\frac{2}{36}$ $\frac{1}{36}$

$\begin{array}{l}\mathrm{Then},\mathrm{expectation}\mathrm{of}\mathrm{X}\\ \mathrm{ }=\mathrm{E}\left(\mathrm{X}\right)\\ \mathrm{ }=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)\\ \mathrm{ }=2×\frac{1}{36}+3×\frac{2}{36}+4×\frac{3}{36}+5×\frac{4}{36}+6×\frac{5}{36}+7×\frac{6}{36}\\ +8×\frac{5}{36}+9×\frac{4}{36}+10×\frac{3}{36}+11×\frac{2}{36}+12×\frac{1}{36}\\ \mathrm{ }=7\\ \mathrm{And}\mathrm{ }\mathrm{E}\left({\mathrm{X}}^{2}\right)= {2}^{2}×\frac{1}{36}+{3}^{2}×\frac{2}{36}+{4}^{2}×\frac{3}{36}+{5}^{2}×\frac{4}{36}+{6}^{2}×\frac{5}{36}+{7}^{2}×\frac{6}{36}\\ +\mathrm{ }{8}^{2}×\frac{5}{36}+{9}^{2}×\frac{4}{36}+{\left(10\right)}^{2}×\frac{3}{36}+11×\frac{2}{36}+{\left(12\right)}^{2}×\frac{1}{36}\\ \mathrm{ }=\frac{329}{6}=54.833\\ \mathrm{Thus},\\ \mathrm{Var}\left(\mathrm{X}\right)=\mathrm{E}\left({\mathrm{X}}^{2}\right)-{\left(\mathrm{E}\left(\mathrm{X}\right)\right)}^{2}\\ \mathrm{ }=54.833-{\left(7\right)}^{2}\\ \mathrm{ }=54.833-49\\ \mathrm{ }=5.833\\ \mathrm{Standard}\mathrm{deviation}\\ \mathrm{ }\mathrm{\sigma }=\sqrt{5.833}\\ \mathrm{ }=2.415\end{array}$

Q.14 A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Ans

$\begin{array}{l}\mathrm{There}\mathrm{are}15\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{class}.\mathrm{Each}\mathrm{student}\mathrm{has}\mathrm{the}\mathrm{same}\\ \mathrm{chance}\mathrm{to}\mathrm{be}\mathrm{chosen}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{each}\mathrm{student}\mathrm{to}\mathrm{be}\mathrm{selected}=\frac{1}{15}\\ \mathrm{The}\mathrm{frequency}\mathrm{table}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{data}\mathrm{is}\mathrm{as}\mathrm{follows}:\end{array}$

 X 14 15 16 17 18 19 20 21 f 2 1 2 3 1 2 3 1

$\begin{array}{l}P\left(X=14\right)=\frac{2}{15}\\ P\left(X=15\right)=\frac{1}{15}\\ P\left(X=16\right)=\frac{2}{15}\\ P\left(X=17\right)=\frac{3}{15}\\ P\left(X=18\right)=\frac{1}{15}\\ P\left(X=19\right)=\frac{2}{15}\\ P\left(X=20\right)=\frac{3}{15}\\ P\left(X=21\right)=\frac{1}{15}\\ Thus,\text{the probability distribution of X is}\end{array}$

 X 14 15 16 17 18 19 20 21 P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$

$\begin{array}{l}Then,\text{â€‹}\text{mean of X}=E\left(X\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=14×\frac{2}{15}+15×\frac{1}{15}+16×\frac{2}{15}+17×\frac{3}{15}+18×\frac{1}{15}\\ \text{}\text{}\text{}\text{}+19×\frac{2}{15}+20×\frac{3}{15}+21×\frac{1}{15}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{15}\left(28+15+32+51+18+38+60+21\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{263}{15}=17.53\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}E\left({X}^{2}\right)={14}^{2}×\frac{2}{15}+{15}^{2}×\frac{1}{15}+{16}^{2}×\frac{2}{15}+{17}^{2}×\frac{3}{15}+{18}^{2}×\frac{1}{15}\\ \text{}\text{}\text{}\text{}+{19}^{2}×\frac{2}{15}+{20}^{2}×\frac{3}{15}+{21}^{2}×\frac{1}{15}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4683}{15}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=312.2\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Variance\left(X\right)=E\left({X}^{2}\right)-{\left[E\left(X\right)\right]}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=312.2-{\left(17.53\right)}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=312.2-307.4177\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.7823\approx 4.78\\ \text{Standard deviation}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{Variance\left(X\right)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{4.78}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2.19\end{array}$

Q.15 In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X).

Ans

$\begin{array}{l}\mathrm{P}\left(\mathrm{X}=0\right)=30\mathrm{%}=0.3\\ \mathrm{P}\left(\mathrm{X}=1\right)=70\mathrm{%}=0.7\\ \mathrm{So},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 0 1 P(X) 0.3 0.7

$\begin{array}{l}Then,\text{\hspace{0.17em}}E\left(X\right)=0×0.3+1×0.7\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}E\left({X}^{2}\right)={0}^{2}×0.3+{1}^{2}×0.7\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Var\left(X\right)=E\left({X}^{2}\right)-{\left[E\left(X\right)\right]}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7-{\left(0.7\right)}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.7-0.49\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.21\end{array}$

Q.16 The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1 (B) 2 (C) 5 (D) 8/3

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{be}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{representing}\mathrm{a}\mathrm{number}\mathrm{on}\mathrm{the}\mathrm{die}\mathrm{.}\\ \mathrm{The}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{observations}\mathrm{is}\mathrm{six}\mathrm{.}\\ \mathrm{S}=\left\{1, 1, 1, 2, 2, 5\right\}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\frac{3}{6}=\frac{1}{2}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\frac{2}{6}=\frac{1}{3}\\ \mathrm{P}\left(\mathrm{X}=5\right)=\frac{1}{6}\\ \mathrm{So},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 1 2 5 P (X) $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{6}$

$\begin{array}{l}E\left(X\right)=1×\frac{1}{2}+2×\frac{1}{3}+5×\frac{1}{6}\\ \text{}=\frac{3+4+5}{6}\\ \text{}=\frac{12}{6}=2\\ Therefore,\text{corect option is B}\text{.}\end{array}$

Q.17 Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
(A) 37/221 (B) 5/13 (C) 1/13 (D) 2/13

Ans

$\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{in}\mathrm{deck}=52\\ \mathrm{Number}\mathrm{of}\mathrm{aces}=\mathrm{4}\\ \mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{aces}\mathrm{obtained}.\mathrm{Therefore},\mathrm{X}\mathrm{can}\\ \mathrm{take}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{values}\mathrm{of}0,1,\mathrm{or}2\mathrm{.}\\ \mathrm{In}\mathrm{a}\mathrm{deck}\mathrm{of}52\mathrm{cards}, 4\mathrm{cards}\mathrm{are}\mathrm{aces}.\mathrm{Therefore},\mathrm{there}\mathrm{are}\\ 48\mathrm{non}–\mathrm{ace}\mathrm{cards}\mathrm{.}\\ \therefore \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{P}\left(0\mathrm{ace}\mathrm{card}\mathrm{and}2\mathrm{non}–\mathrm{ace}\mathrm{cards}\right)\\ =\frac{\mathrm{C}_{0}^{4}×\mathrm{C}_{2}^{48}}{\mathrm{C}_{2}^{52}}\\ =\frac{1128}{1326}\\ \mathrm{P}\left(\mathrm{X}=1\right)=\mathrm{P}\left(1\mathrm{ace}\mathrm{card}\mathrm{and}1\mathrm{non}–\mathrm{ace}\mathrm{cards}\right)\\ =\frac{\mathrm{C}_{1}^{4}×\mathrm{C}_{1}^{48}}{\mathrm{C}_{2}^{52}}\\ =\frac{192}{1326}\\ \mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{P}\left(2\mathrm{ace}\mathrm{card}\mathrm{and}0\mathrm{non}–\mathrm{ace}\mathrm{cards}\right)\\ =\frac{\mathrm{C}_{2}^{4}×\mathrm{C}_{0}^{48}}{\mathrm{C}_{2}^{52}}\\ =\frac{6}{1326}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{distribution}\mathrm{of}\mathrm{X}\mathrm{is}\end{array}$

 X 0 1 2 P(X) $\frac{1128}{1326}$ $\frac{192}{1326}$ $\frac{6}{1326}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\therefore E\left(X\right)=0×\frac{1128}{1326}+1×\frac{192}{1326}+2×\frac{6}{1326}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{192}{1326}+\frac{12}{1326}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{204}{1326}=\frac{2}{13}\\ Therefore,\text{the correct option is D}\text{.}\end{array}$

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