Probability

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

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

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

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

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

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

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

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

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) * P (H) = 3/4 x 3/4 = 9/16

P (X = 1) = P (one tail) = P (HT) + P (TH)

= 3/4 . 1/4 + 1/4 . 3/4

= 3/16 + 3/16= 3/8

P (X = 2) = P (two tails) = P (TT) = 1/4 x 1/4 = 1/16

Therefore, the required probability distribution is as follows.

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) * P (H) = 3/4 x 3/4 = 9/16

P (X = 1) = P (one tail) = P (HT) + P (TH)

= 3/4 . 1/4 + 1/4 . 3/4

= 3/16 + 3/16= 3/8

P (X = 2) = P (two tails) = P (TT) = 1/4 x 1/4 = 1/16

Therefore, the required probability distribution is as follows.

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-defective and 1 defective) ${=}^4{C}_1.\left(\frac{1}{5}\right).{\left(\frac{4}{5}\right)}^3=\frac{256}{625}$

$P\left(X=2\right)=P$ (2 non-defective and 2 defective) ${=}^4{C}_2.{\left(\frac{1}{5}\right)}^2.{\left(\frac{4}{5}\right)}^2=\frac{96}{625}$

$P\left(X=3\right)=P$ (1 non-defective and 3 defective) ${=}^4{C}_3.{\left(\frac{1}{5}\right)}^3.\left(\frac{4}{5}\right)=\frac{16}{625}$

$P\left(X=4\right)=P$ (0 non-defective and 4 defective) ${=}^4{C}_4.{\left(\frac{1}{5}\right)}^4.{\left(\frac{4}{5}\right)}^0=\frac{1}{625}$

Therefore, the required probability distribution is as follows.