Probability

81. Total number of balls in the urn = 25

Balls bearing mark 'X' = 10

Balls bearing mark 'Y' = 15

p = P (ball bearing mark 'X') =10/25 = 2/5

q = P (ball bearing mark 'Y') =15/25 = 3/5

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with 'Y' mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p =2/5.

$\therefore P=\left(Z=z\right)={\mathrm{nC}}_z$
$p^{n-z}{q}^z$

P (all will bear 'X' mark) $=P\left(Z=0\right)={\mathrm{6C}}_0$
$\mathrm{}{\left(\frac{2}{5}\right)}^6={\left(\frac{2}{5}\right)}^6$

P (not more than 2 bear 'Y' mark) $=P\left(Z\le 2\right)$

$=P\left(Z=0\right)+P\left(Z=1\right)+P\left(Z=2\right)\\ ={\mathrm{6C}}_0$
$\mathrm{}{\left(p\right)}^6{\left(q\right)}^0+6C_1$
$\mathrm{}{\left(p\right)}^5{\left(q\right)}^1+{\mathrm{6C}}_2$
$\begin{array}{l}\mathrm{}{\left(p\right)}^4{\left(q\right)}^2\\ ={\left(\frac{2}{5}\right)}^6+6{\left(\frac{2}{5}\right)}^5\left(\frac{3}{5}\right)+15{\left(\frac{2}{5}\right)}^4{\left(\frac{3}{5}\right)}^2\\ ={\left(\frac{2}{5}\right)}^4\left[{\left(\frac{2}{5}\right)}^2+6\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+15{\left(\frac{3}{5}\right)}^2\right]\\ ={\left(\frac{2}{5}\right)}^4\left[\frac{4}{25}+\frac{36}{25}+\frac{135}{25}\right]\\ ={\left(\frac{2}{5}\right)}^4\left[\frac{175}{25}\right]\\ =7{\left(\frac{2}{5}\right)}^4\end{array}$

P (at least one ball bears 'Y' mark) $=P\left(Z\ge 1\right)=1-P\left(Z=0\right)$

$=1-{\left(\frac{2}{5}\right)}^6$

P (equal number of balls with 'X' mark and 'Y' mark) $=P\left(Z=3\right)$

$={\mathrm{6C}}_3$
$\begin{array}{l}\mathrm{}{\left(\frac{2}{5}\right)}^3{\left(\frac{3}{5}\right)}^3\\ =\frac{20*8*27}{15625}\\ =\frac{864}{3125}\end{array}$

80. It is given that $90\mathrm{\%}$ of people are right handed.

$p=P$ (right handed) $=90\mathrm{\%}=\frac{90}{100}=\frac{9}{10}$ and $q=P$ (left handed) $=1-\frac{9}{10}=\frac{1}{10}$

Using Binomial distribution,

Probability more than $6$ people are right handed is given by

$=\sum _{r=7}^{10}{\mathrm{10c}}_r$

${\displaystyle \mathrm{}{p}^r{q}^{n-r}}$

$=\sum _{r=7}^{10}{\mathrm{10c}}_r$

${\displaystyle \mathrm{}{\left(\frac{9}{10}\right)}^r{\left(\frac{1}{10}\right)}^{10-r}}$

Hence, probability of having at most $6$ right handed people:

$=1-P$ (more than $6$ people are right handed)

$=1-\sum _{r=7}^{10}{\mathrm{10c}}_r$

${\displaystyle \mathrm{}{\left(\frac{9}{10}\right)}^r{\left(\frac{1}{10}\right)}^{10-r}}$

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}*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 with n=5 and $P=\frac{1}{3}$

$\therefore P=\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x\\ ={\mathrm{5C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{2}{3}\right)}^{5-x}.{\left(\frac{1}{3}\right)}^x\end{array}$

P (guessing more than 4 correct answers) $=P\left(X\ge 4\right)$

$=P\left(X=4\right)+P\left(X=5\right)\\ ={\mathrm{5C}}_4$
$\mathrm{}\left(\frac{2}{3}\right).{\left(\frac{1}{3}\right)}^4+{\mathrm{5C}}_5$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{3}\right)}^5\\ =5.\frac{2}{3}.\frac{1}{81}+1.\frac{1}{243}\\ =\frac{10}{243}+\frac{1}{243}\\ =\frac{11}{243}\end{array}$

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 distribution with n=20 and $P=\frac{1}{2}$

$\therefore P=\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}.p^x,$ where $x=0,1,2,...n$

$={\mathrm{20C}}_x$

$\mathrm{}{\left(\frac{1}{2}\right)}^{20-x}.{\left(\frac{1}{2}\right)}^x\\ ={\mathrm{20C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{2}\right)}^{20}\end{array}$

P (at least 12 questions answered correctly) $=P\left(X\ge 12\right)$

$=P\left(X=12\right)+P\left(X=13\right)+...+P\left(X=20\right)\\ ={\mathrm{20C}}_{12}$

$\mathrm{}{\left(\frac{1}{2}\right)}^{20}+{\mathrm{20C}}_{13}$
$\mathrm{}{\left(\frac{1}{2}\right)}^{20}+...+{\mathrm{20C}}_{20}$
$\mathrm{}{\left(\frac{1}{2}\right)}^{20}\\ ={\left(\frac{1}{2}\right)}^{20}.[20C_{12}$
$\mathrm{}+{\mathrm{20C}}_{13}$
$\mathrm{}+...+{\mathrm{20C}}_{20}$
$\begin{array}{l}\mathrm{}]\end{array}$

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)

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