Chapter 6.3, Problem 112E

(a)

To determine

To find: The expected number of spins to win

Answer to Problem 112E

The expected number of spins to win is 38.

Explanation of Solution

Given information:

Number of spins $=\text{ 15}$

Probability of the ball landing in any slot $=\frac{1}{38}$ .

Concept used:

If the trials are repeated until a success occurs and the probability of success is same in each trial, then it is the case of geometric distribution.

In this case, a bet is placed until a win and the probability of winning in each chance is same and was independent. So, this follows geometric distribution.

Calculation:

Let X be the number of spins for a success

So, $\text{X }~\text{ Geometric with p = }\left(\frac{1}{38}\right)$

  $P(X=k)={{\displaystyle (1-p)}}^{k-1}\times p$

Mean of X

  $\text{E }\left(\text{X}\right)=\frac{1}{p}=\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$38$}\right.}=38$

(b)

To determine

To check: whether it is probable to won in less than or equal to 3.

Answer to Problem 112E

It is surprising if Martin won in 30 or fewer spins.

Explanation of Solution

Calculation:

Probability of winning in 3 or fewer spins

  $P(X\le 3)=0+{{\displaystyle \left(1-\frac{1}{38}\right)}}^{1-1}\times \frac{1}{38}+{{\displaystyle \left(1-\frac{1}{38}\right)}}^{2-1}\times \frac{1}{38}+{{\displaystyle \left(1-\frac{1}{38}\right)}}^{3-1}\times \frac{1}{38}$

  $P\left(X\le 3\right)=0.07689$

It can be seen that the probability of 3 or fewer is too low.

So, it is surprising if Martin won in 30or fewer spins.