Answer: Option D) $(5 / 6)^{5}-(2 / 3)^{5}$
Approach $1$ :
When a dice is rolled $5$ times, the largest number rolled is $5$ can happen only when in none of the throws $6$ appeared and $5$ appeared at least one time.
Total probability of $6$ not appearing in any throw = $(5/6)^5$
But among these cases, there are cases where $5$ and $6$ never appears. Eg: $(1,2,3,4,2)$
Hence we need to subtract the cases where only the $4$ no.s, i.e. $(1,2,3,4)$ appear.
The probability of $5$ and $6$ never appearing = $(4/6)^5$
Thus, the actual probability = $(5/6)^5 – (4/6)^5 = (5 / 6)^{5}-(2 / 3)^{5}$ [Option D]
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Approach $2$ :-
We can fill the five throws like :
Case $1$: ($5$ appears once, and in rest $4$ throws, any no. b/w $(1,2,3,4)$ appear) = $5C1*4^4$
Case $2$: ($5$ appears twice, and in rest $3$ throws, any no. b/w $(1,2,3,4)$ appear) = $5C2*4^3$
.
.
Case $5$: ($5$ appears five times, and in rest $0$ throws, any no. b/w $(1,2,3,4)$ appear) = $5C1*4^0$
Thus the number of favorable cases are:-
$5C1*4^4 + 5C2*4^3+ 5C3*4^2+5C4*4^1+5C5*4^0 = 1280+640+160+20+1 = 2101$
Total no. of cases = $6^5 = 7776$ (any of the $6$ no.s fall in the $5$ throws)
Thus actual prob = fav cases/tot cases = $2101/7776$
If we see Option D) $(5 / 6)^{5}-(2 / 3)^{5} = (3125-1024)/7776 = 2101/7776$ Answer.
