91Ó°ÊÓ

$A=$ rolling the dice with $1$ die.

$B=$rolling the $1$with all $5$dice.

$C=$rolling a yahtzee.

Dice are fair, and since the dice have a total of $6$possible outcomes, rolling a $1$is a decent choice.

$P\left(A\right)=\frac{1}{6}$

To answer the problem, the multiplication rule is utilised.

$P\left(B\right)=P\left(A\right)×P\left(A\right)×……×P\left(A\right)\left\{5\text{time repetitions}\right\}$$=\left[P\right(A){]}^5$

$={\left[\frac{1}{6}\right]}^5$

When playing Yahtzee, use the same number on all five dice. There are a total of $6$numbers that could be rolled $(1,2,3,4,5,6)$, and each of these numbers has the same chance of being rolled with all $5$dice.

Use the addition rule when two conditions are mutually exclusive.

$P\left(C\right)=6P\left(B\right)$

$=6{\left(\frac{1}{6}\right)}^5$

Given data:

$A=$rolling the dice with $1$die.

$B=$rolling the $1$with all $5$dice.

$C=$rolling a Yahtzee.

In the case of mutually exclusive conditions, use the addition rule.

$P\left(C\right)=6P\left(B\right)$

$=6{\left(\frac{1}{6}\right)}^5$

$=\frac{1}{1296}$

In the case of mutually exclusive conditions, use the addition rule.

$P(X≤25)=P(X=1)+P(X=2)+…P(X=24)+P(X=25)$$=\underset{k=1}{\overset{25}{∑}}{\left(1-\frac{1}{1296}\right)}^{k-1}\left(\frac{1}{1296}\right)$

$=0.0191$

Because the probability is tiny, Nassir should not be startled if he does not acquire a Yahtzee in the first $25$ rolls.