Question

Two persons A and B take turns in throwing a pair of dice. The first person to throw 9 from both dice will be awarded the prize. If A throws first, then the probability that B wins the game is

A. 9/17

B. 8/17

C. 8/9

D. 1/9

Answer 1

Correct option is  B. 8/17

Let’s find out the probability of getting ‘9’ from throw of a pair of dice first: 9 can be obtained in the given 4 cases →

[(3,6), (4,5), (5,4), (6,3)]

Getting 9 → \(\cfrac4{36}=\cfrac19\)

Not getting 9 → \(\cfrac{32}{36}=\cfrac89\)

B will win when A won’t get 9 in his/her throw and B will get 9 in his/her throw, i.e. B wins after 1 round of throws, P = \(\cfrac89\times\cfrac19\)

Now then,

P(B wins) = P(Getting 9 in 2^nd throw) + P(Getting 9 in 4^th throw) + P(Getting 9 in 6^th throw) + . . .