Two players play the following game : $A$ writes $3, 5, 6$ on three different cards; $B$ writes $8, 9, 10$ on three different cards. Both draw randomly two cards from their collections. Then $A$ computes the product of two numbers he/she has drawn, and $B$ computes the sum of two numbers he/she has drawn. The player getting the larger number wins. What is the probability that $A$ wins?

\frac{1}{3}

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\frac{5}{9}

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\frac{4}{9}

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\frac{1}{9}

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Solution

The correct option is B $\frac{5}{9}$
If $B$ has to win $A$ can either draw $3, 5$ or $3, 6$

If $A$ draw $3, 6$ then $B$ can only draw $9, 10$

Then Probabilty $P = \frac{1}{3} . \frac{1}{3} = \frac{1}{9}$

If $A$ draws $3, 5$ then $B$ can draw any pair of two cards from the collections

Then $P = \frac{1}{3} . \frac{3}{3} = \frac{1}{3}$

Total probability that B wins $= \frac{1}{3} + \frac{1}{9} = \frac{4}{9}$
Thus, the probability of A wins $= 1 - \frac{4}{9} = \frac{5}{9}$
Option B is correct.

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