Chapter PIV, Problem 33RE

(a)

To determine

To find: that for the first time on what roll a 6 will be entitled to get.

Answer to Problem 33RE

6

Explanation of Solution

Probability of occurring 6

Suppose X is number of times roll to get first 6

X is geom(6)

Number of rolls expected

  $\frac{1}{p}=\frac{1}{\frac{1}{6}}=6$

It can be expected to roll 6 times for getting first 6

(b)

To determine

To find: the expected score on the basis of provided information.

Answer to Problem 33RE

4.17

Explanation of Solution

Given:

Rolled number of times is 6-1 =5.

Calculation:

Suppose X get any number other than 6.

  $P\left(X\right)=\frac{5}{6}$

Now $X_1+X_2+X_3+X_4+X_5$ is the total score

  $\begin{array}{c}E\left(X_1+X_2+X_3+X_4+X_5\right)=E\left(X_1\right)+E\left(X_2\right)+E\left(X_3\right)+E\left(X_4\right)+E\left(X_5\right)\\ =5\left(\frac{5}{6}\right)\\ =\frac{25}{6}\\ =4.17\end{array}$

Expected score is 4.17.

(c)

To determine

To find: the probability that, without having a 6, will roll that many times.

Answer to Problem 33RE

5/6

Explanation of Solution

Given:

Rolled number of times rolled is 6.

Calculation:

Probability of not getting $6=\frac{5}{6}$

X is binom $\left(6,\frac{5}{6}\right)$

  $\begin{array}{c}P\left(X=6\right)=C{}_6{\left(\frac{5}{6}\right)}^6{\left(\frac{1}{6}\right)}^0\\ =0.33\end{array}$

Therefore, there is 33% possibility that of not getting 6 is $\frac{5}{6}$ .