Probability Ques 71
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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Answer:
Correct Answer: 71.$(\frac{2}{5})$
Solution:
Formula:
- $5$ 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} $
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