Probability

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) ×

54. It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

$\therefore P\left(X=0\right)=P$ (4 non-defective and 0 defective) ${=}^4{C}_0.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

$P\left(X=1\right)=P$ (3 non-d

53. When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

(i) Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 X 4/6 = 4/9

P (X = 1)

(i) When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads.

∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

Therefore, X can take the value of 0, 1, or 2.

It is known that,

P (HH) = P (HT) = P (TH) = P (TT) = 1/4

P (X = 0) = P (TT) = 1/4

P (X = 1) =

51. A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T) = |6 - 0| = 6

X (5 H, 1 T)  = |5 - 1| = 4

X (4 H, 2 T)  = |4 - 2| = 2

X (3 H, 3 T)  = 3 - 3| = 0

X (2 H, 4 T)  = |2 - 4| = 2

X (1 H, 5 T)  =

50. There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.