Matrix Matching : A bag contains some white and some black balls, all combinations being equally likely. The total number of balls in the bag is 12. Four balls are drawn at random from the bag at random without replacement

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$\frac{14}{33}$

$\frac{1}{5}$

$\frac{70}{429}$

$\frac{13}{165}$

Probability that all the four balls are black is equal to

If the bag contains 10 black and 2 white balls then the probability that all four balls are black is equal to

If all the four balls are black, then the probability that the bag contains 10 black balls is equal to

Probability that two balls are black and two are white is

answer is , , A.

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Detailed Solution

Let $E_{i}$ be the event in the bag contains $i$ black and $(12 - i)$ white balls $(i = 0, 1, 2, \dots, 12)$ and A denotes the event that the four balls drawn all are black, then

$P (E_{i}) = \frac{1}{13} (i = 0, 1, 2, \dots, 12)$

$P (A / E_{i}) = 0 for i = 0, 1, 2, 3$

And $P (A / E_{i}) = \frac{^{i} C_{4}}{^{12} C_{4}} for 4 \leq i \leq 12$ then

(A): $P (A) = \sum_{i = 0}^{12} P (E_{i}) . P (A / E_{i})$

$\begin{array}{l} = \frac{1}{13} \times \frac{1}{^{12} C_{4}} [^{4} C_{4} +^{5} C_{4} + \dots +^{12} C_{4}] \\ = \frac{^{13} C_{5}}{13 \times^{12} C_{4}} = \frac{1}{5} \end{array}$

(B): $P (A / E_{10}) = \frac{^{10} C_{4}}{^{12} C_{4}} = \frac{14}{33}$

(C):By Bayers theorem

$\begin{array}{l} P (E_{10} / A) = \frac{P (E_{10}) . P (A / E_{10})}{P (A)} \\ = \frac{\frac{1}{13} \times \frac{14}{33}}{\frac{1}{5}} \\ = \frac{70}{429} \end{array}$

(D): Let B denote the probability of getting 2 white and 2 black balls

Then $P (\frac{B}{E_{i}}) = 0 for i = 0, 1 or 11,12$

$P (\frac{B}{E_{i}}) = \frac{^{i} C_{2} \times 12 -^{i} C_{2}}{^{12} C_{4}} for 2 \leq i \leq 10$

$\begin{array}{l} P (B) = \sum_{i = 0}^{12} P (E_{i}) P (\frac{B}{E_{i}}) \\ = \frac{1}{13} \times \frac{1}{^{12} C_{4}} [2 (^{2} C_{2} \times^{10} C_{2} +^{3} C_{2} \times^{9} C_{2} + \dots +^{10} C_{2} \times^{2} C_{2})] \\ = \frac{1}{13} \times \frac{1}{^{12} C_{4}} [2 (^{2} C_{2} \times^{10} C_{2} +^{3} C_{2} \times^{9} C_{2} + \dots +^{5} C_{2} \times^{7} C_{2}) +^{6} C_{2} \times^{6} C_{2}] \\ = \frac{1}{13} \times \frac{1}{495} (1287) \\ = \frac{1}{5} \end{array}$