Chapter 7, Problem 118CR

The longest “run” of S’s in the sequence SSFSSSSFFS has length 4, corresponding to the S’s on the fourth, fifth, sixth, and seventh positions. Consider a binomial experiment with n = 4, and let y be the length in the longest run of S’s.

1. a. When p = 0.5, the 16 possible outcomes are equally likely. Determine the probability distribution of y in this case (first list all outcomes and the y value for each one). Then calculate μ_y.
2. b. Repeat Part (a) for the case p = 0.6.
3. c. Let z denote the longest run of either S’s or F’s. Determine the probability distribution of z when p = 0.5.

To determine

Find the probability distribution of y when p = 0.5.

Obtain ${\mu }_y$.

Answer to Problem 118CR

The probability distribution of y is obtained as given below:

y	0	1	2	3	4
p(y)	0.0625	0.4375	0.3125	0.1250	0.0625

The mean for the random variable y is 1.6875.

Explanation of Solution

Calculation:

It is given that, the longest “run” of S’s in the sequence SSFSSSSFFS has length 4. This length corresponds to S’s on the fourth, fifth, sixth, and seventh positions.

Define the random variable y as the length in the longest run of S’s. Here, the random variable y follows binomial distribution with n = 4.

When p = 0.5, there are 16 possible outcomes and these outcomes are equally likely to occur.

One of the possible outcomes is SSSS. Here, longest run of S is 4. Hence, the random variable y takes the value 4. The corresponding probability is, $\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$. Similarly, the other outcomes and its corresponding y value and probability are listed in the below table.

Outcome	y	Probability
SSSS	4	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SSSF	3	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SSFS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SSFF	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFSS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFSF	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFFS	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFFF	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSSS	3	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSSF	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSFS	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSFF	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFSS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFSF	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFFS	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFFF	0	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$

The probability distribution of y is obtained as given below:

y	0	1	2	3	4
p(y)	0.0625	$\begin{array}{l}0.0625\times 7\\ =\text{0}\text{.4375}\end{array}$	$\begin{array}{l}0.0625\times 5\\ =\text{0}\text{.3125}\end{array}$	$\begin{array}{l}0.0625\times 2\\ =\text{0}\text{.1250}\end{array}$	$\begin{array}{l}0.0625\times 1\\ =\text{0}.0625\end{array}$

Mean:

The mean for the random variable y is obtained as given below:

y	p(y)	$y\cdot p\left(y\right)$
0	0.0625	0
1	0.4375	0.4375
2	0.3125	0.625
3	0.125	0.375
4	0.0625	0.25
		${\displaystyle \sum _{y=1,\mathrm{...},4}y\cdot p\left(y\right)}=1.6875$

From the above table, the mean is,

$\begin{array}{c}{\mu }_y={\displaystyle \sum _{y=1,\mathrm{...},4}y\cdot p\left(y\right)}\\ =1.6875\end{array}$

Thus, the mean for the random variable y is 1.6875.

To determine

Find the probability distribution of y when p = 0.6.

Obtain ${\mu }_y$.

Answer to Problem 118CR

The probability distribution of y is obtained as given below:

y	0	1	2	3	4
p(y)	0.0256	0.3264	0.3456	0.1728	0.1296

The mean for the random variable y is 2.0544.

Explanation of Solution

Calculation:

Define the random variable y as the length in the longest run of S’s.

When p = 0.6, there are 16 possible outcomes and these outcomes are equally likely to occur.

One of the possible outcomes is SSSS. Here, S occurs for 4 times. Hence, the random variable y takes the value 4. The corresponding probability is, $\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.6\right)=\text{0}\text{.1296}$. Similarly, the other outcomes and its corresponding y value and probability are listed in the below table.

Outcome	y	Probability
SSSS	4	$\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.6\right)=\text{0}\text{.1296}$
SSSF	3	$\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.4\right)=\text{0}\text{.0864}$
SSFS	2	$\left(0.6\right)\left(0.6\right)\left(0.4\right)\left(0.6\right)=\text{0}\text{.0864}$
SSFF	2	$\left(0.6\right)\left(0.6\right)\left(0.4\right)\left(0.4\right)=\text{0}\text{.0576}$
SFSS	2	$\left(0.6\right)\left(0.4\right)\left(0.6\right)\left(0.6\right)=\text{0}\text{.0864}$
SFSF	1	$\left(0.6\right)\left(0.4\right)\left(0.6\right)\left(0.4\right)=\text{0}\text{.0576}$
SFFS	1	$\left(0.6\right)\left(0.4\right)\left(0.4\right)\left(0.6\right)=\text{0}\text{.0576}$
SFFF	1	$\left(0.6\right)\left(0.4\right)\left(0.4\right)\left(0.4\right)=\text{0}\text{.0384}$
FSSS	3	$\left(0.4\right)\left(0.6\right)\left(0.6\right)\left(0.6\right)=\text{0}\text{.0864}$
FSSF	2	$\left(0.4\right)\left(0.6\right)\left(0.6\right)\left(0.4\right)=\text{0}\text{.0576}$
FSFS	1	$\left(0.4\right)\left(0.6\right)\left(0.4\right)\left(0.6\right)=\text{0}\text{.0576}$
FSFF	1	$\left(0.4\right)\left(0.6\right)\left(0.4\right)\left(0.4\right)=\text{0}\text{.0384}$
FFSS	2	$\left(0.4\right)\left(0.4\right)\left(0.6\right)\left(0.6\right)=\text{0}\text{.0576}$
FFSF	1	$\left(0.4\right)\left(0.4\right)\left(0.6\right)\left(0.4\right)=\text{0}\text{.0384}$
FFFS	1	$\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.6\right)=\text{0}\text{.0384}$
FFFF	0	$\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.4\right)=\text{0}\text{.0256}$

The probability distribution of y is obtained as given below:

y	p(y)
0	0.0256
1	$0.0567+0.0567+0.0384+0.0576+0.0384+0.0384+0.0384=\text{0}.3264$
2	$0.0864+0.0567+0.0864+0.0576+0.0576=\text{0}.3456$
3	$0.0864+0.0864=0.1728$
4	0.1296

Mean:

The mean for the random variable y is obtained as given below:

y	p(y)	$y\cdot p\left(y\right)$
0	0.0256	0
1	0.3264	0.3264
2	0.3456	0.6912
3	0.1728	0.5184
4	0.1296	0.5184
		${\displaystyle \sum _{y=1,\mathrm{...},4}y\cdot p\left(y\right)}=2.0544$

From the above table, the mean is,

$\begin{array}{c}{\mu }_y={\displaystyle \sum _{y=1,\mathrm{...},4}y\cdot p\left(y\right)}\\ =2.0544\end{array}$

Thus, the mean for the random variable y is 2.0544.

To determine

Find the probability distribution of z when p = 0.5.

Answer to Problem 118CR

The probability distribution of z is obtained as given below:

z	1	2	3	4
p(z)	0.1250	0.5000	0.2500	0.1250

Explanation of Solution

Calculation:

Define the random variable z as the length in the longest run either S’s or F’s.

When p = 0.5, there are 16 possible outcomes and these outcomes are equally likely to occur.

One of the possible outcomes is SSSS. Here, S occurs for 4 times. Hence, the random variable y takes the value 4. The corresponding probability is, $\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$. Similarly, the other outcomes and its corresponding y value and probability are listed in the below table.

Outcome	z	Probability
SSSS	4	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SSSF	3	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SSFS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SSFF	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFSS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFSF	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFFS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
SFFF	3	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSSS	3	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSSF	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSFS	1	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FSFF	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFSS	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFSF	2	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFFS	3	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$
FFFF	4	$\left(0.5\right)\left(0.5\right)\left(0.5\right)\left(0.5\right)=\text{0}\text{.0625}$

The probability distribution of z is obtained as given below:

z	1	2	3	4
p(z)	$\begin{array}{l}0.0625\times 2\\ =\text{0}\text{.125}0\end{array}$	$\begin{array}{l}0.0625\times 8\\ =\text{0}\text{.5}\end{array}$	$\begin{array}{l}0.0625\times 4\\ =\text{0}\text{.250}\end{array}$	$\begin{array}{l}0.0625\times 2\\ =\text{0}\text{.125}\end{array}$