A die is thrown again and again until three 5s' are obtained. Find the probability of obtaining the third 5 in the seventh throw of the die.

$\frac{3125}{93312}$

The correct answer is Option (3) → $\frac{3125}{93312}$

Let E be the event of getting 5, then

$p = P(E)=\frac{1}{6}$, so $q = 1-\frac{1}{6}=\frac{5}{6}$.

As die is rolled seven times, so there are 7 Bernoullian trials i.e. $n = 7$.

Thus, we have a binomial distribution with $p =\frac{1}{6},q=\frac{5}{6}, n = 7$.

Required probability = P(third 5 on seventh throw of die)

= P(two 5s in first six throws) × P(a 5 on seventh throw)

$={^6C}_2(\frac{1}{6})^2(\frac{5}{6})^2×\frac{1}{6}$

$=15×\frac{1}{36}×\frac{625}{1296}×\frac{1}{6}=\frac{3125}{93312}$