Yahtzee $(5.3,6.3)$In the game of Yahtzee, $5$six-sided dice are rolled simultaneously. To get a Yahtzee, the player must get the same number on all $5$dice.

(a) Luis says that the probability of getting a Yahtzee in one roll of the dice is ${\left(\frac{1}{6}\right)}^5$. Explain why Luis is wrong.

(b) Nassir decides to keep rolling all 5 dice until he gets a Yahtzee. He is surprised when he still hasn’t gotten a Yahtzee after$25$rolls. Should he be? Calculate an appropriate probability to support your answer .

$5$ six sides dice have been rolled simultaneously.

To obtain Yahtzee, all $5$ dice must have same numbers.

According to Luis, the probability of getting a Yahtzee is ${\left(\frac{1}{6}\right)}^5$.

The following numbers could be obtained by players $1,2,3,4,5$and $6$.

Here all the dice are independent.

Thus, the probability of getting a Yahtzee is calculated as:

$P\left(\text{Yahtzee}\right)=6\times {\left(\frac{1}{6}\right)}^5$

$5$six sides dice have been rolled simultaneously.

To obtain Yahtzee, all $5$dice must have same numbers.

According to Luis, the probability of getting a Yahtzee is localid="1650542498223" ${\left(\frac{1}{6}\right)}^5$.

From above part, it is known that

$$\begin{gathered} P\left(\text{Yahtzee}\right)=6\times {\left(\frac{1}{6}\right)}^5 \\ =0.0007716 \end{gathered}$$

$$\begin{gathered} P\left(\text{No Yahtzee}\right)=1-6\times {\left(\frac{1}{6}\right)}^5 \\ =0.9992284 \end{gathered}$$

The probability of not getting Yahtzee even after $25$rolls is computed as:

localid="1650622029853" $$\begin{gathered} P(\text{No Yahtzee}{)}^{25}=(0.9992254{)}^{25} \\ =0.9809 \\ =98.09\% \end{gathered}$$

Since, the probability is above $0.05$. Thus, it could be concluded that the probability of not getting Yahtzee is likely. Thus, Nasir should not be surprised.