First slide

Theorems of probability

Question

The probabilities of four cricketers A, B, C and D scoring more than 50 runs in a match are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{10}$ . It is known that exactly two of the players scored more than 50 runs in a particular match. The probability that these players were A and B is

Difficult

A

$\frac{27}{65}$
B

$\frac{5}{6}$
C

$\frac{1}{6}$
D

None

Solution

Let $E_{1}$ be the event that exactly two players scored more than 50 runs then $P (E_{1}) = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} \times \frac{9}{10}$ $+ \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} \times \frac{9}{10} + \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{1}{10} + \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{9}{10}$ $+ \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{1}{10} + \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} \times \frac{1}{10} = \frac{65}{240}$
Let $E_{2}$ be the event A and B scored more than 50 runs, then $P (E_{1} \cap E_{2}) = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} \times \frac{9}{10} = \frac{27}{240}$

$∴$ desired probability = $P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{27}{65}$

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