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. The probability that A wins is

Question

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. The probability that A wins is (A) (1/3) (B) (5/9) (C) (4/9) (D) (1/9)

Tardigrade Admin · Accepted Answer

For A to win, A can drawn either 3,6 or 5,6 . If A draws 3,6 then B can draw only 8 and 9 . Probability

= (\frac{1}{3}) (\frac{1}{3}) = \frac{1}{9}

.
Also, if A draws 5, 6 then B can draw, any two probability

= \frac{1}{3} \times 1 = \frac{1}{3}

So, required probability

= \frac{1}{9} + \frac{1}{3} = \frac{4}{9}

.

$\frac{1}{3}$

$\frac{5}{9}$

$\frac{4}{9}$

$\frac{1}{9}$

31​

95​

94​

91​

For A to win, A can drawn either 3,6 or 5,6 . If A draws 3,6 then B can draw only 8 and 9 . Probability =(31​)(31​)=91​. Also, if A draws 5, 6 then B can draw, any two probability =31​×1=31​ So, required probability =91​+31​=94​.

$\frac{1}{3}$

$\frac{5}{9}$

$\frac{4}{9}$

$\frac{1}{9}$