Ncert Solutions Maths class 12th

86. Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability (p) of getting a head at the toss of a coin is 1/2.

∴ p = 1/2 ⇒ q = 1/2

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

It is given that,

P (getting at least one head) > 90/100

P (x ≥ 1) > 0.9

⇒ 1 − P (x = 0) > 0.9

$1-{\mathrm{nC}}_0$
$.\frac{1}{2^n}>0.9\\ {\mathrm{nC}}_0$
$\begin{array}{l}.\frac{1}{2^n}<0.1\\ \frac{1}{2^n}<0.1\\ 2^n>\frac{1}{0.1}\\ 2^n>10......(1)\end{array}$

The minimum value of n that satisfies the given inequality is 4.

Thus, the man should toss the coin 4 or more than 4 times.

85. The probability of success is twice the probability of failure.

Let the probability of failure be x.

∴ Probability of success = 2x

$\begin{array}{l}x+2x=1\\ \Rightarrow 3x=1\\ \Rightarrow x=\frac{1}{3}\\ \therefore 2x=\frac{2}{3}\end{array}$

Let $p=\frac{1}{3}$ and $q=\frac{2}{3}$

Let X be the random variable that represents the number of successes in six trials.

By binomial distribution, we obtain

$P\left(X=x\right)={\mathrm{nC}}_x$
$p^{n-x}{q}^x$

Probability of at least 4 successes $=P\left(X\ge 4\right)$

$=P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)\\ ={\mathrm{6C}}_4$
$\mathrm{}{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2+{\mathrm{6C}}_5$
$\mathrm{}{\left(\frac{2}{3}\right)}^5\left(\frac{1}{3}\right)+{\mathrm{6C}}_6$
$\begin{array}{l}\mathrm{}{\left(\frac{2}{3}\right)}^6\\ =\frac{15{\left(2\right)}^4}{3^6}+\frac{6{\left(2\right)}^5}{3^6}+\frac{{\left(2\right)}^6}{3^6}\\ =\frac{{\left(2\right)}^4}{{\left(3\right)}^6}\left[15+12+4\right]\\ =\frac{31*2^4}{{\left(3\right)}^6}\\ =\frac{31}{9}{\left(\frac{2}{3}\right)}^4\end{array}$

84. A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday, then the year will have 53 Tuesdays.

Number of total days in a week = 7

Number of favourable days = 2

Therefore, P (the year will have 53 Tuesday) = 2/7

83. The probability of getting a six in a throw of die is 1/6 and not getting a six is 5/6.

Let p = 1/6 and q = 5/6

The probability that the 2 sixes come in the first five throws of the die is

${\mathrm{5C}}_2$
$\mathrm{}{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^3=\frac{10*{\left(5\right)}^3}{{\left(6\right)}^5}$

∴ Probability that third six comes in the sixth throw =

$\begin{array}{l}=\frac{10*{\left(5\right)}^3}{{\left(6\right)}^5}*\frac{1}{6}\\ =\frac{10*125}{{\left(6\right)}^6}\\ =\frac{10*125}{46656}\\ =\frac{625}{23328}\end{array}$

82. Let p and q respectively be the probabilities that the player will clear and knock down the hurdle.

$\begin{array}{l}\therefore P=\frac{5}{6}\\ \Rightarrow q=1-p=1-\frac{5}{6}=\frac{1}{6}\end{array}$

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Therefore, by binomial distribution, we obtain

$P\left(X=x\right)={\mathrm{nC}}_x$
$p^{n-x}{q}^x$

P (player knocking down less than 2 hurdles) $=P\left(X<2\right)$

$=P\left(X=0\right)+P\left(X=1\right)\\ ={\mathrm{10C}}_0$

$\mathrm{}{\left(q\right)}^0{\left(p\right)}^{10}+{\mathrm{10C}}_1$
$\begin{array}{l}\mathrm{}\left(q\right){\left(p\right)}^9\\ ={\left(\frac{5}{6}\right)}^{10}+10.\frac{1}{6}.{\left(\frac{5}{6}\right)}^9\\ ={\left(\frac{5}{6}\right)}^9\left[\frac{5}{6}+\frac{10}{6}\right]\\ =\frac{5}{2}{\left(\frac{5}{6}\right)}^9\\ =\frac{{\left(5\right)}^{10}}{2*{\left(6\right)}^9}\end{array}$

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}$