Question

A die is continually rolled until the total sum of all rolls exceeds 300. Approximate the probability that at least 80 rolls are necessary.

Short answer

$P\{Y_{79}\le 300\}=0.9429$

Step-by-step solution

Step 1. Given information

Let X_idenote the value of the die, of i^throll.

Step 2. Mean and variance of Xi

Mean = $E\left(X_i\right)=\frac{(1+2+3+4+5+6)}{6}=\frac{21}{6}=\frac{7}{2}$

Variance =role="math" localid="1648052296820" $V\left(X_i\right)=\frac{35}{12}$

Step 3. Introducing Yn

Assuming that die is continually rolled until the total sum of all rolls exceeds 300.

Let Y_nrepresents the total sum of all rolls with n number of rolls.

$Y_n=\underset{1}{\overset{n}{\sum X_i}}$

Step 4. Mean and variance of Yn

Mean = $E\left(Y_n\right)=n\times E\left(X_i\right)=\frac{7n}{2}$

Variance =$V\left(Y_n\right)=n\times V\left(X_i\right)=\frac{35n}{12}$

Step 5. To find

The probability that at least 80 rolls are necessary can be written as -

$P\left\{\underset{1}{\overset{79}{\sum X_i\le 300}}\right\}=P\left\{Y_{79}\le 300\right\}$

Step 6. Using central limit theorem

$$\begin{gathered} P\{Y_{79}\le 300\}=P\left\{\frac{Y_{79}-{\displaystyle \frac{7\left(79\right)}{2}}}{\sqrt{{\displaystyle \frac{35\left(79\right)}{12}}}}\le \frac{300-{\displaystyle \frac{7\left(79\right)}{2}}}{\sqrt{{\displaystyle \frac{35\left(79\right)}{12}}}}\right\} \\ P\{Y_{79}\le 300\}=P\left\{Z\le 1.58\right\} \\ P\{Y_{79}\le 300\}=0.9429 \end{gathered}$$

Step 7. Final answer

The probability that at least 80 rolls are necessary is 0.9429