On a toss of two dice, A throws a total of 5. Then the probability that he

Question

On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is

Options:

A . 245

B . 25

C . 181

D . 19

Answer: Option B
:
B
let P(A) be the probability of throwing total of 5 and P(B) the probability of throwing total 7.

∴ P (A) = \frac{4}{36} = \frac{1}{9}, P (B) = \frac{6}{36} = \frac{1}{6}, P (A o r B) = \frac{5}{18} a n d P (A a n d B) = 0 P (n e i t h e r A n o r B) = \frac{13}{18}, ∴ P (5 b e f o r e 7) = \frac{1}{9} + \frac{13}{18} . \frac{1}{9} + \frac{13}{18} . \frac{13}{18} . \frac{1}{9} + . . . . \infty = \frac{1}{9} [\frac{1}{1 - \frac{13}{18}}] = \frac{2}{5} .

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