Question

1 point) Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 318. Approximate the probability that at least 84 rolls are needed to get this sum.

Answer 1

Answer:

The probability that at least 84 rolls are needed to get a sum of 318 is 0.9616.

Step-by-step explanation:

A six-sided fair dice is rolled.

Let X = outcome of rolling a 6-sided dice.

The probability of getting any of the 6 faces is, p = 1/6.

The expected value of X is:

$E(X)=\sum x. P(X=x)\\=(1\times \frac{1}{6})+(2\times \frac{1}{6})+(3\times \frac{1}{6})+(4\times \frac{1}{6})+(5\times \frac{1}{6})+(6\times \frac{1}{6})\\=3.5$

Compute the variance of X as follows:

$V(X)=E(X^{2})-[E(X)]^{2}\\=[(1^{2}\times \frac{1}{6})+(2^{2}\times \frac{1}{6})+...+(6^{2}\times \frac{1}{6})]-(3.5)^{2}\\=2.92$

As the dice is rolled a lot of times, the distribution of ∑X can be approximated by a Normal distribution.

Then the probability that at least 84 rolls are needed to get a sum of 318 is same as the probability that the sum in 83 trials is less than 318.

The mean of the distribution of ∑X is:

$\mu_{\sum X}=n\times E(X)=83\times 3.5=290.5$

The variance of the distribution of ∑X is:

$\sigma^{2}_{\sum X}=n\times V(X)=83\times 2.92=242.36$

Thus, $\sum X\sim N(290.5, 242.36)$

Compute the value of P (∑X < 83) as follows:

$P(\sum X<318)=P(\frac{\sum X-\mu_{\sum X}}{\sigma_{\sum X}}<\frac{318-290.5}{\sqrt{242.36}})\\=P(Z<1.77)\\=0.9616$

*Use a z-table for the probability.

Thus, the probability that at least 84 rolls are needed to get a sum of 318 is 0.9616.