Question

A pair of dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is

• A. 1/5
• B. 2/5
• C. 3/5
• D. 4/5

Answer 1

Answer: (B)

2/5

Answer 2

5 can be thrown in 4 ways and 7 can be thrown in 6 ways in a single throw of two dice.

Number of ways of throwing neither 5 nor 7 = 36 – (4 + 6)

= 26

Probability of throwing a sum of 5 in a throw = $\frac { 4 } { 36 } = \frac { 1 } { 9 }$

Probability of throwing neither 5 nor 7 = $\frac { 26 } { 36 } = \frac { 13 } { 18 }$

Required probability = $\frac { 1 } { 9 } + \frac { 13 } { 18 } \left( \frac { 1 } { 9 } \right) + \left( \frac { 13 } { 18 } \right) ^ { 2 } \frac { 1 } { 9 } + \ldots$

= $\frac { 1 / 9 } { 1 - 13 / 18 } = \frac { 2 } { 5 }$