First slide

Random variable and probability distribution

Question

Two persons A and B get together once a week to play game. They always play 4 games. From past experience Mr. A wins 2 of the 4 games just as often as he wins 3 of the 4 games. If the probability of winning a game by Mr. A is (Given the probability of A’s winning a game is a non-zero constant less than one) p, then value of 10p is

Difficult

Solution

$\begin{matrix} ^{4} C_{2} \cdot p^{2} \cdot {(1 - p)}^{2} =^{4} C_{3} \cdot p^{3} \cdot (1 - p) \\ 6 (1 - p) = 4 p \\ 4 p + 6 p = 6 \\ 10 p = 6 \\ p = \frac{6}{10} \\ = 0.6 \end{matrix}$

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