JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Probability

The experiment stops at the fifth throw only if the fourth and fifth throws are both fours, these being the first pair of consecutive fours. That forces the third throw not to be a four, since otherwise a pair would already have closed at the fourth throw, and the first two throws must not themselves be two consecutive fours. So the fifth and fourth throws are fixed as fours, each with probability $\frac{1}{6}$; the third throw is anything but four, probability $\frac{5}{6}$; and the first two throws can be any pair except four-four. There are $6^2=36$ possibilities for the first two throws, of which only four-four is forbidden, leaving $35$. Multiplying, the favourable outcomes number $35\times 5=175$ out of $6^5$, so the probability is $\frac{175}{6^5}$.