Probability

79. Given, Men having grey hair $=5\mathrm{\%}$

Women having grey hair $=0.25\mathrm{\%}$

Total people with grey hair $=(5+0.25)\mathrm{\%}$ $=5.25\mathrm{\%}$

Probability that the selected person’s hair is of male $=\frac{5}{5.25}=\frac{1}{1.05}=\frac{100}{105}=\frac{20}{21}$

78. i. Here, Sample space of given condition,

$S=\{(B,B),(B,G),(G,B),(G,G)\}$

Let, $E:$ denotes the event both children are male

$F:$ denotes the event at least one of the children is a male

$\therefore E\cap F=\{(B,B)\}$

$\Rightarrow P(E\cap F)=\frac{1}{4}$

Here, $P(E)=\frac{1}{4}$ and $P(F)=\frac{3}{4}$

$\Rightarrow P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{1}{3}$

ii. Let, $A:$ event that both children are female

$B:$ event that elder child is female

$A=\{G,G\}$ $P(A)=\frac{1}{4}$

$B=\{(G,G),(G,B)\}$ $P(B)=\frac{2}{4}=\frac{1}{2}$

$A\cap B=\{(G,G)\}$ $P(A\cap B)=\frac{1}{4}$

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{4}\times 2=\frac{1}{2}$

$$77. $A$ is a subset of $B$

$\Rightarrow A\cap B=A$

$\therefore P(A\cap B)=P(B\cap A)=P(A)$

$=P(B/A)=\frac{P(B\cap A)}{P(A)}=1$

$A\cap B\ne 0$

$(A\cap B)=0$

$\Rightarrow P(B/A)=\frac{P(B\cap A)}{P(A)}=0$

76. The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.

Probability of getting 5 in a single throw of the die, p = 1/6

$\therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}$

Clearly, X has the probability distribution with n=7 and $P=\frac{1}{6}$

$\therefore P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x={\mathrm{7C}}_x$
$\mathrm{}{\left(\frac{5}{6}\right)}^{7-x}.{\left(\frac{1}{6}\right)}^x$

P (getting 5 exactly twice) $=P\left(X=2\right)$

$={\mathrm{7C}}_2$
$\begin{array}{l}\mathrm{}{\left(\frac{5}{6}\right)}^5.{\left(\frac{1}{6}\right)}^2\\ =21.{\left(\frac{5}{6}\right)}^5.\frac{1}{36}\\ =\left(\frac{7}{12}\right)+{\left(\frac{5}{6}\right)}^5\end{array}$

74. The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.

Probability of getting a correct answer is, p = 1/3

$\therefore q=1-p=1-\frac{1}{3}=\frac{2}{3}$

Clearly, X has a binomial distribution wi

73.  X is the random variable whose binomial distribution is B(6,1/2).

Therefore, n = 6 and p = ½

$\therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}\\ Then,P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x\\ ={\mathrm{6C}}_x$
$\mathrm{}{\left(\frac{1}{2}\right)}^{6-x}.{\left(\frac{1}{2}\right)}^x\\ ={\mathrm{6C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{2}\right)}^6\end{array}$

It can be seen that P (X=x) will be maximum, if ${\mathrm{6C}}_x$
 will be maximum.

$Then,{\mathrm{6C}}_0$
$\mathrm{}={\mathrm{6C}}_6$
$\mathrm{}=\frac{6!}{0!6!}=1\\ {\mathrm{6C}}_1$
$\mathrm{}={\mathrm{6C}}_5$
$\mathrm{}=\frac{6!}{1!5!}=6\\ {\mathrm{6C}}_2$
$\mathrm{}={\mathrm{6C}}_4$
$\mathrm{}=\frac{6!}{2!4!}=15\\ {\mathrm{6C}}_3$
$\begin{array}{l}\mathrm{}=\frac{6!}{3!3!}=20\end{array}$

The value of ${\mathrm{6C}}_3$
 is maximum. Therefore, for x=3, P(X=x) is maximum.

Therefore, P(X=3) is maximum.

72. Let X represent the number of correctly answered questions out of 20 questions.

The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.

$\begin{array}{l}\therefore P=\frac{1}{2}\\ \therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}\end{array}$

X has a binomial dist

71. Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

Since the balls are drawn with replacement, the trials are Bernoulli trials.

X has a binomial distribution with n = 4 and p = 1/10

$\therefore q=1-p=1-\frac{1}{10}=\frac{9}{10}\\ \therefore P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}.p^x,x=1,2,...n\\ ={\mathrm{4C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{9}{10}\right)}^{4-x}.{\left(\frac{1}{10}\right)}^x\end{array}$

P (none marked with 0)=P (X=0)

$={\mathrm{4C}}_0$
$\begin{array}{l}\mathrm{}{\left(\frac{9}{10}\right)}^4.{\left(\frac{1}{10}\right)}^0\\ =1.{\left(\frac{9}{10}\right)}^4\\ ={\left(\frac{9}{10}\right)}^4\end{array}$

70. Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, p = 0.05

$\therefore q=1-p=1-0.05=0.95$

X has a binomial distribution with n=5 and $p=0.05$

$={\mathrm{5C}}_x$
$\mathrm{}{\left(0.95\right)}^{5-x}.{\left(0.05\right)}^x$

$P\left(none\right)=P\left(X=0\right)$

$={\mathrm{5C}}_0$
$\begin{array}{l}\mathrm{}{\left(0.95\right)}^5.{\left(0.05\right)}^0\\ =1\times {\left(0.95\right)}^5\\ ={\left(0.95\right)}^5\end{array}$

(ii) P (not more than one) $=P\left(X\le 1\right)$

$=P\left(X=0\right)+P\left(X=1\right)\\ =5C_0$
$\mathrm{}{\left(0.95\right)}^5\times {\left(0.05\right)}^0+{\mathrm{5C}}_1$
$\begin{array}{l}\mathrm{}{\left(0.95\right)}^4\times {\left(0.05\right)}^1\\ =1\times {\left(0.95\right)}^5+5\times {\left(0.95\right)}^4\times {\left(0.05\right)}^1\\ ={\left(0.95\right)}^5+\left(0.05\right){\left(0.95\right)}^4\\ ={\left(0.95\right)}^4\times 1.2\end{array}$

(iii) P (more than 1) $=P\left(X>1\right)$

$=1-P\left(X\le 1\right)$

$=1-P$ (not more th

69. Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.

In a well shuffled deck of 52 cards, there are 13 spade cards.

$\begin{array}{l}\Rightarrow p=\frac{13}{52}=\frac{1}{4}\\ \therefore q=1-\frac{1}{4}=\frac{3}{4}\end{array}$

X has a binomial distribution with n=5 and $p=\frac{1}{4}$

$={\mathrm{5C}}_x$
$\mathrm{}{\left(\frac{3}{4}\right)}^{5-x}{\left(\frac{1}{4}\right)}^x$

P (all five cards are spades)&nb