Ncert Solutions Maths class 12th

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.

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

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

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

62. When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) +

61. The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) $=\frac{2}{30}=\frac{1}{15}$

F

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 d

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

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

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

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

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

Therefore, the required probability distribution is as follows.

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

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

58. (a) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒ k =1/6

(b) P (X < 2) = P (X = 0) + P (X = 1)

= k + 2k

= 3k

=3/6 = 1/2

$\begin{array}{l}P\left(X\le 2\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\\ =k+2k+3k\\ =6k\\ =\frac{6}{6}\\ =1\end{array}$

$\begin{array}{l}P\left(X\ge 2\right)=P\left(X=2\right)+P\left(X>2\right)\\ =3k+0\\ =3k\\ =\frac{3}{6}\\ =\frac{1}{2}\end{array}$

57.  (i) Since, the sum of all the probabilities of a distribution is 1.

$\begin{array}{l}\Rightarrow 0+k+2k+2k+3k+k^2+2k^2+\left(7k^2+k\right)=1\\ \Rightarrow 10k^2+9k=0\\ \Rightarrow \left(10k-1\right)\left(k+1\right)=0\\ \Rightarrow k=-1,\frac{1}{4}\end{array}$

K=-1 is not possible as the probability of an event is never negative.

$\therefore k=\frac{1}{10}$

(ii) $P\left(X<3\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)$

$\begin{array}{l}=0+k+2k\\ =3k\\ =3\times \frac{1}{10}\\ =\frac{3}{10}\end{array}$

(iii) $P\left(X>6\right)=P\left(X=7\right)$

$\begin{array}{l}=7k^2+k\\ =7\times {\left(\frac{1}{10}\right)}^2+\frac{1}{10}\\ =\frac{7}{100}+\frac{1}{10}\\ =\frac{17}{100}\end{array}$

(iv) $P\left(0<X<3\right)=P\left(X=1\right)+P\left(X=2\right)$

$\begin{array}{l}=k+2k\\ =3k\\ =3\times \frac{1}{10}\\ =\frac{3}{10}\end{array}$

56. Let the probability of getting a tail in the biased coin be x.

∴ P (T) = x

⇒ P (H) = 3x

For a biased coin, P (T) + P (H) = 1

$\begin{array}{l}\Rightarrow x+3x=1\\ \Rightarrow 4x=1\\ \Rightarrow x=\frac{1}{4}\\ \therefore P\left(T\right)=\frac{1}{4},and,P\left(H\right)=\frac{3}{4}\end{array}$

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

Let X be the random variable representing the number of tails.

∴ P (X = 0) = P (no tail) = P (H) ×

55. Let the probability of getting a tail in the biased coin be x.

∴ P (T) = x

⇒ P (H) = 3x

For a biased coin, P (T) + P (H) = 1

$\begin{array}{l}\Rightarrow x+3x=1\\ \Rightarrow 4x=1\\ \Rightarrow x=\frac{1}{4}\\ \therefore P\left(T\right)=\frac{1}{4},and,P\left(H\right)=\frac{3}{4}\end{array}$

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

Let X be the random variable representing the number of tails.

∴ P (X = 0) = P (no tail) = P (H) ×