Q60 / 90 · KVPY 2011 (SB/SX) · Mathematics
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?
- A$\frac{1}{3}$
- B$\frac{5}{9}$
- Correct answerC$\frac{4}{9}$
- D$\frac{1}{9}$
Solution
For A to win, A can draw either 3,6 or 5,6 . If A draws 3,6 then B can draw only $8 \& 9$
$\text { Prob. }=\frac{1}{3} \cdot \frac{1}{3}=\frac{1}{9}$
If A draws 5,6 then $\mathrm{B}$ can draw, any two
$\begin{array}{l}
\text { Probability }=\frac{1}{3} \cdot 1=\frac{1}{3} \\
\text { Probability }=\frac{1}{9}+\frac{1}{3}=\frac{4}{9}
\end{array}$
Hence, the correct answer is option: 'C'.