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 the two numbers he/she has drawn, and B computes the sum of the two numbers he/she has drawn. The player getting the larger number wins. What is the probability that A wins?
$
  {\text{A}}{\text{. }}\dfrac{1}{3} \\
  {\text{B}}{\text{. }}\dfrac{5}{9} \\
  {\text{C}}{\text{. }}\dfrac{4}{9} \\
  {\text{D}}{\text{. }}\dfrac{1}{9} \\
$

Answer 1

Hint: We have to find the probability of winning A, and as given A computes the product of numbers written on two cards drawn so we have to write all possibilities of A and then for B write the sum of numbers written on cards in all possibilities.

Complete step-by-step answer:

Now we have
Possibilities of A of selecting two cards
A can get: (3,5) (6,5) (3,6)
And hence products of all possibilities are following:
Possible products: {15 , 30 , 18}
Possibility of B of selecting two cards are following:
B can get: (8,9) (9,10) (8,10)
Possible sum obtained from all possibilities is following:
Possible sums: {17 , 19 , 18}
For A to win:
When B gets (8,9)
A should get (6,5) (3,6) :$\dfrac{1}{3} \times \dfrac{1}{3} + \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{2}{9}$
When B gets (9,10)
A should get (6,5) : $\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$
When B gets (8,10)
A should get (6,5) : $\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$
Total probability for A to win:
$\dfrac{2}{9} + \dfrac{1}{9} + \dfrac{1}{9} = \dfrac{4}{9}$
So, option C is correct.

Note: In this question, the winner would be the one who gets the larger number. Since there is a probability that both A and B can get (3,6) and (8,10) making their product and sum as 18. Hence, there is a $\dfrac{1}{9}$ probability of getting a draw.