Chapter 6, Problem 23P

1. A roulette wheel has alternating red and black numbered slots into which the ball finally stops to determine the winner. If a gambler always bets on black to win, then
2. What is the probability of winning at least 40 times in a series of 64 spins? (Note that at lea.st 40 wins means 40 or more.)
3. What is the probability of winning more than 40 times in a series of 64 spins?
4. Based on your answers to a and b, what is the
6. probability of winning exactly 40 times?

To determine

Find the required probability.

Answer to Problem 23P

Solution:

1. The probability of winning at least 40 times in a series of 64 spins is 0.0301.
2. The probability of winning more than 40 times in a series of 64 spins is 0.0166.
3. The probability of winning exactly 40 times in a series of 64 spins is 0.0135.

Explanation of Solution

When normal distribution is used to approximate binomial distribution, for $P\left(X\ge m\right)$, use $P\left(X>m-0.5\right)$. Therefore, use $P\left(X>39.5\right)$ in place of $P\left(X>40\right)$. For $P\left(z\le 1.88\right)$, look into standard normal table at $z=1.88$ .

When normal distribution is used to approximate binomial distribution, for $P\left(X>m\right)$, use $P\left(X>m+0.5\right)$. Therefore, use $P\left(X>40.5\right)$ in place of $P\left(X>40\right)$. For $P\left(z\le 2.13\right)$, look into standard normal table at $z=2.13$ .

To get the probability of winning exactly 40 times, subtract probability of winning more than 40 times from the probability of winning at least 40 times (40 or more) in a series of 64 spins.

Given:

Number of questions, $n=64$ , and probability of winning is 0.5.

Formula used:

$\begin{array}{l}q=1-p\\ \mu =pn\\ \sigma =\sqrt{npq}\\ z=\frac{x-\mu }{\sigma }\end{array}$

Calculation:

1. Each slot has 2 color choices (red/black). Therefore, the probability of guessing correctly for any individual question is:

   $\begin{array}{c}p=\frac{\text{correct choice}}{\text{total choices}}\\ =\frac{1}{2}\\ =0.5\end{array}$

   The probability of guessing correctly for any individual question is:

   $\begin{array}{c}q=1-p\\ =1-0.5\\ =0.5\end{array}$

   The distribution of number of wins will form a normal shaped distribution with mean and standard deviation as shown below:

   $\begin{array}{c}\mu =pn\\ =0.5\left(64\right)\\ =32\\ \sigma =\sqrt{npq}\\ =\sqrt{64\left(0.5\right)\left(0.5\right)}\\ =4\end{array}$

   We are looking for the probability of winning at least 40 times in a series of 64 spins, so we will use lower real limit for 40, which is 39.5. The z score for score $x=39.5$ is:

   $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{39.5-32}{4}\\ =1.88\end{array}$

   The probability of winning at least 40 times in a series of 64 spins is:

   $\begin{array}{c}P\left(X\ge 40\right)=P\left(X>39.5\right)\\ =P\left(z>1.88\right)\\ =1-P\left(z\le 1.88\right)\\ =1-0.9699\\ =0.0301\end{array}$

2. We are looking for the probability of winning more than 40 times in a series of 64 spins, so we will use upper real limit for 40, which is 40.5. The z score for score $x=40.5$ is:

   $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{40.5-32}{4}\\ =2.13\end{array}$

   The probability of winning more than 40 times in a series of 64 spins is:

   $\begin{array}{c}P\left(X>40\right)=P\left(X>40.5\right)\\ =P\left(z>2.13\right)\\ =1-P\left(z\le 2.13\right)\\ =1-0.9834\\ =0.0166\end{array}$

3. By using results of part (a) and part (b), the probability of winning exactly 40 times in a series of 64 spins is:

   $\begin{array}{c}P\left(X>40\right)=P\left(X\ge 40\right)-P\left(X>40\right)\\ =0.0301-0.0166\\ =0.0135\end{array}$

Conclusion:

The probability of winning at least 40 times in a series of 64 spins is 0.0301.

The probability of winning more than 40 times in a series of 64 spins is 0.0166.

The probability of winning exactly 40 times in a series of 64 spins is 0.0135.