Ncert Solutions Maths class 12th

67. The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

$\begin{array}{l}p=\frac{6}{36}=\frac{1}{6}\\ \therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}\end{array}$

Clearly, X has the binomial distribution with n=4, $p=\frac{1}{6}$ , and $q=\frac{5}{6}$

$\therefore P\left(X=x\right){=}^n{C}_x{q}^{n-x}{p}^x$ , where $x=0,1,2,3...n$

$=4C_x$
$\mathrm{}{\left(\frac{5}{6}\right)}^{4-x}.{\left(\frac{1}{6}\right)}^x\\ =4C_x$
$\begin{array}{l}\mathrm{}.{\frac{5}{6^4}}^{4-x}\end{array}$

$\therefore P$ (2 successes) $=P\left(X=2\right)$

$={\mathrm{4C}}_2$
$\begin{array}{l}\mathrm{}.{\frac{5}{6^4}}^{4-2}\\ =6.\frac{25}{1296}\\ =\frac{25}{216}\end{array}$

66. The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is, p - 3/6 = 1/2

q = 1 - p = 1/2

X has a binomial distribution.

P (at most 5 successes) $=P\left(X\le 5\right)$

$=1-P\left(X>5\right)\\ =1-P\left(X=6\right)\\ =1-{\mathrm{6C}}_6$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{2}\right)}^6\\ =1-\frac{1}{64}\\ =\frac{63}{64}\end{array}$

66. Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

X	0	1	2
P (X)	1128/1326	192/1326	6/1326

$Then,E\left(X\right)={\displaystyle \sum _{}^{}{P}_i}{X}_i$

$\begin{array}{l}=0*\frac{1128}{1326}+1*\frac{192}{1326}+2*\frac{6}{1326}\\ =\frac{204}{1326}\\ =\frac{2}{13}\end{array}$

Therefore, option D is correct

65. Let X be the random variable representing a number on the die.

The total number of observations is six.

$\begin{array}{l}\therefore P\left(X=1\right)=\frac{3}{6}=\frac{1}{2}\\ P\left(X=2\right)=\frac{2}{6}=\frac{1}{3}\\ P\left(X=5\right)=\frac{1}{6}\end{array}$

Therefore, the probability distribution is as follows.

$Mean=E\left(X\right)={\displaystyle \sum _{}^{}{P}_i}{X}_i$

$\begin{array}{l}=\frac{1}{2}*1+\frac{1}{3}*2+\frac{1}{6}*5\\ =\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\\ =\frac{3+4+5}{6}\\ =\frac{12}{6}\\ =2\end{array}$

Therefore, option (B) is correct.

64. It is given that P (X = 0) = 30% = 30/100 = 0.3

P (X = 1) = 70% = 70/100 = 0.7

Therefore, the probability distribution is as follows.

$\begin{array}{l}Then,E\left(X\right)={\displaystyle \sum _{}^{}{X}_{i}P}\left(X_i\right)\\ =0*0.3+1*0.7\\ =0.7\end{array}$

$\begin{array}{l}E\left(X^2\right)={\displaystyle \sum _{}^{}{X^2}_{i}P}\left(X_i\right)\\ =0^2*0.3+{\left(1\right)}^2*0.7\\ =0.7\end{array}$

It is know that,  $Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2$

= 0.7 − (0.7)²

= 0.7 − 0.49

= 0.21

63. There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18) =115, P(X = 19) =2/15, P(X = 20) =3/15, P(X = 21) =1/15

Therefore, the probability distribution of random variable X is as follows.

Then, mean of X = E(X)

$\begin{array}{l}={\displaystyle \sum _{}^{}{X}_{i}P}\left(X_i\right)\\ =14*\frac{2}{15}+15*\frac{1}{15}+16*\frac{2}{15}+17*\frac{3}{15}+18*\frac{1}{15}+19*\frac{2}{15}+20*\frac{3}{15}+21*\frac{1}{15}\\ =\frac{1}{15}\left(28+15+32+51+18+38+60+21\right)\\ =\frac{263}{15}\\ =17.53\end{array}$

$\begin{array}{l}E\left(X^2\right)={\displaystyle \sum _{}^{}{X^2}_{i}P}\left(X_i\right)\\ ={\left(14\right)}^2.\frac{2}{15}+{\left(15\right)}^2.\frac{1}{15}+{\left(16\right)}^2.\frac{2}{15}+{\left(17\right)}^2.\frac{3}{15}+{\left(18\right)}^2.\frac{1}{15}+{\left(19\right)}^2.\frac{2}{15}+{\left(20\right)}^2.\frac{3}{15}+{\left(21\right)}^2.\frac{1}{15}\\ =\frac{1}{15}.\left(392+225+512+867+324+722+1200+441\right)\\ =\frac{4683}{15}\\ =312.2\end{array}$

$\begin{array}{l}\therefore Variance\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2\\ =312.2-{\left(\frac{263}{15}\right)}^2\\ =312.2-307.4177\\ =4.7823\\ \approx 4.78\end{array}$

Standard derivation $\begin{array}{l}=\surd variance\; (X)\\ =\surd 4.78\\ =2.186\approx 2.19\end{array}$

60. Two dice thrown simultaneously is the same the die thrown 2 times.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) = 25/36

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

=2 (1/6 x 5/6) = 10/36

P (X = 2) = P (six on both the dice) =1/36

Therefore, the required probability distribution is as follows.

Then, expectation of X = E (X) =∑ X iP (Xi)

= 0 x 25/36 + 1 x 10/36 + 2 x 1/36

= 1/3

59. Let X denote the success of getting heads.

Therefore, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0, 1, 2, or 3

$\begin{array}{l}\therefore P\left(X=0\right)=P\left(TTT\right)\\ =P\left(T\right).P\left(T\right).P\left(T\right)\\ =\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\\ =\frac{1}{8}\end{array}$

$\begin{array}{l}\therefore P\left(X=1\right)=P\left(HHT\right)+P\left(HTH\right)+P\left(THH\right)\\ =\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\\ =\frac{3}{8}\end{array}$

$\begin{array}{l}\therefore P\left(X=2\right)=P\left(HHT\right)+P\left(HTH\right)+P\left(THH\right)\\ =\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\\ =\frac{3}{8}\end{array}$

$\begin{array}{l}\therefore P\left(X=3\right)=P\left(HHH\right)\\ =\frac{1}{2}*\frac{1}{2}*\frac{1}{2}\\ =\frac{1}{8}\end{array}$

Therefore, the required probability distribution is as follows.

Mean of $XE\left(X\right),\mu ={\displaystyle \sum _{}^{}{X}_{i}P\left(X_i\right)}$

$\begin{array}{l}=0*\frac{1}{8}+1*\frac{3}{8}+2*\frac{3}{8}+3*\frac{1}{8}\\ =\frac{3}{8}+\frac{3}{4}+\frac{3}{8}\\ =\frac{3}{2}\\ =1.5\end{array}$