Chapter 14, Problem 14.26E

a.

To determine

To obtain: The probability of winning with a bet on red in a single play of roulette.

Answer to Problem 14.26E

The probability of winning a bet on red colored is 0.47.

Explanation of Solution

Given info:

The roulette wheel consists of 38 slots numbered as 0, 00 and 1 to 36 in which 0,00 are green colored, 18 are red colored and 18 are black.

Calculation:

Here, the sample size is $n=38$, number of red colored is 18

The probability of winning a bet on red in a single play of roulette is,

$\begin{array}{c}P\left(\text{Winning}\text{a}\text{bet}\text{on}\text{red}\text{colored}\right)=\frac{\text{Number}\text{of}\text{red}\text{colored}}{\text{Sample}\text{size}}\\ =\frac{18}{38}\\ =0.4737\end{array}$

Thus, the probability of winning a bet on red colored is 0.47.

b.

To determine

To obtain: The distribution of random variable X.

Answer to Problem 14.26E

The distribution of X is binomial.

X	0	1	2	3	4
$P\left(X=x\right)$	0.0789	0.2799	0.3723	0.2201	0.0488

Explanation of Solution

Given info:

A bet is placed on red, every time when the roulette is played for four times.

Calculation:

Define the random variable X as the number of win, when bet is placed on red every time.

Also, there are two possible outcomes (winning the bet on red or losing the bet on red) and the probability of success is the probability that winning when placing bet on red each time (p) is 0.47 and not winning, when placing bet on red each time is 0.53 $\left(=1-0.47\right)$. Thus, X follows the binomial distribution.

Therefore, winning when placing bet on red every time follows the binomial distribution with sample size $\left(n\right)$ of 4 and the probability $\left(p\right)$ with 0.47.

Thus, the value of n and p, if X has a binomial distribution is 4 and 0.47 respectively.

The probability value when $X=0$:

The binomial distribution formula is,

$P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^x{q}^{n-x}$

Substitute $n=4$, p as 0.47 and q as $0.53\left(=1-0.47\right)$.

$\begin{array}{c}P\left(X=0\right)=\frac{4!}{0!\left(4-0\right)!}{\left(0.47\right)}^0{\left(0.53\right)}^{4-0}\\ =1\times 1\times 0.0789\\ =0.0789\end{array}$

Thus, the probability value with $X=0$ is 0.0789.

The probability value when $X=1$:

Substitute $n=4$, p as 0.47 and q as $0.53\left(=1-0.47\right)$.

$\begin{array}{c}P\left(X=1\right)=\frac{4!}{1!\left(4-1\right)!}{\left(0.47\right)}^1{\left(0.53\right)}^{4-1}\\ =4\times 0.47\times 0.1489\\ =0.2799\end{array}$

Thus, the probability value with $X=1$ is 0.2799.

The probability value when $X=2$:

Substitute $n=4$, p as 0.47 and q as $0.53\left(=1-0.47\right)$.

$\begin{array}{c}P\left(X=2\right)=\frac{4!}{2!\left(4-2\right)!}{\left(0.47\right)}^2{\left(0.53\right)}^{4-2}\\ =6\times 0.2209\times 0.2809\\ =0.3723\end{array}$

Thus, the probability value with $X=2$ is 0.3723.

The probability value with $X=3$:

Substitute $n=4$, p as 0.47 and q as $0.53\left(=1-0.47\right)$.

$\begin{array}{c}P\left(X=3\right)=\frac{4!}{3!\left(4-3\right)!}{\left(0.47\right)}^3{\left(0.53\right)}^{4-3}\\ =4\times 0.1038\times 0.53\\ =0.2201\end{array}$

Thus, the probability value with $X=3$ is 0.2201.

The probability value with $X=4$:

Substitute $n=4$, p as 0.47 and q as $0.53\left(=1-0.47\right)$.

$\begin{array}{c}P\left(X=4\right)=\frac{4!}{4!\left(4-4\right)!}{\left(0.47\right)}^4{\left(0.53\right)}^{4-4}\\ =1\times 0.0488\times 1\\ =0.0488\end{array}$

Thus, the probability value with $X=4$ is 0.0488.

Thus, the probability distribution of X is given below:

X	0	1	2	3	4
$P\left(X=x\right)$	0.0789	0.2799	0.3723	0.2201	0.0488

c.

To determine

To find: The probability of break even.

Answer to Problem 14.26E

The probability of break even is 0.5.

Explanation of Solution

Given info:

Break even means when same amount of bet is placed on every play and win exactly two plays out of four plays.

Calculation:

$\begin{array}{c}P\left(\text{Break}\text{even}\right)=P\left(\text{Winning}\text{exactly}\text{two}\text{play}\text{out}\text{of}\text{four}\text{plays}\right)\\ =\frac{2}{4}\\ =0.5\end{array}$

Thus, the probability of break even is 0.5.

d.

To determine

To find: The probability of losing money.

Answer to Problem 14.26E

The probability of losing money is 0.3125.

Explanation of Solution

Given info:

In four plays, fewer than two are won then money will be lost.

Calculation:

Define the random variable Y “number of times the game is lost”

From part (c) the probability of losing, p is 0.5 thus, q is $0.5\left(=1-0.5\right)$

The probability value for $Y<2$:

The binomial distribution formula is,

$P\left(Y=y\right)=\frac{n!}{\left(n-y\right)!y!}{p}^y{q}^{n-y}$

Substitute $n=4$, and q as $0.5\left(=1-0.5\right)$

$\begin{array}{c}P\left(Y<2\right)=P\left(Y=0\right)+P\left(Y=1\right)\\ =\left(\frac{4!}{0!\left(4-0\right)!}{\left(0.5\right)}^0{\left(0.5\right)}^{4-0}\right)+\left(\frac{4!}{1!\left(4-1\right)!}{\left(0.5\right)}^1{\left(0.5\right)}^{4-1}\right)\\ =\left(1\times 1\times 0.0625\right)+\left(4\times 0.5\times 0.125\right)\\ =0.0625+0.25\end{array}$

                 $=0.3125$

Thus, the probability of losing money is 0.3125.