Probability 3 Question 31
31. A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then, the probability that 5 comes before 7 , is… .
(1989, 2M)
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Solution:
- 5 can be thrown in 4 ways and 7 can be thrown in 6 ways, hence number of ways of throwing neither 5 nor 7 is $\quad 36-(4+6)=26$
$\therefore$ Probability of throwing a five in a single throw with a pair of dice $=\frac{4}{36}=\frac{1}{9}$ and probability of throwing neither 5 nor $7=\frac{26}{36}=\frac{13}{18}$
Hence, required probability
$$ =\frac{1}{9}+\frac{13}{18} \frac{1}{9}+\frac{13}{18}^{2} \frac{1}{9}+\ldots=\frac{\frac{1}{9}}{1-\frac{13}{18}}=\frac{2}{5} $$
