Chapter 15, Problem 18E

(a)

To determine

To find: the probability that the 1^st two grab are the wrong sizes.

Answer to Problem 18E

0.6316

Explanation of Solution

Given:

Medium = 4

Large = 10

Extra large = 6

Calculation:

Two medium shirts are selected. There are 4 + 10 + 6 = 20 shirts in the big box. Also the selection is done without replacement.

Number of shirts = 20 − 4 =16

The total number of favourable event is,

  $\begin{array}{c}\text{Total cases =}\left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 1}}^{st}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 2}}^{nd}\text{ time}\end{array}\right)\\ =20\times 19\\ =380\end{array}$

The favourable cases for first two shirts that are grabbed wrongly is,

  $\begin{array}{c}\text{Favorablecases=}\left(\begin{array}{l}{1}^{st}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{2}^{ed}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\\ =16\times 15\\ =240\end{array}$

Required probability is

  $\begin{array}{c}P\left(\begin{array}{l}\text{firsttwoare}\\ \text{wrongsizes}\end{array}\right)=\frac{\text{Favorablacase}}{\text{Totalcases}}\\ =\frac{240}{380}\\ =0.6316\end{array}$

(b)

To determine

To find: the probability that 1^st medium shirt that find is the 3^rd one check.

Answer to Problem 18E

0.1404

Explanation of Solution

Given:

Total Shirts = 20 shirt

Not medium =16

Medium = 4

Calculation:

The total number of favourable event is

  $\begin{array}{c}\text{Total cases =}\left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 1}}^{st}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 2}}^{nd}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 3}}^{ed}\text{ time}\end{array}\right)\\ =20\times 19\times 18\\ =6840\end{array}$

The favourable cases for first medium shirt that is get in third check can be estimated as:

  $\begin{array}{c}\text{Favorablecases=}\left(\begin{array}{l}{1}^{st}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{2}^{ed}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{3}^{ed}\text{ shirtis }\\ \text{mediumsize}\end{array}\right)\\ =16\times 15\times 4\\ =960\end{array}$

Required probability is

  $\begin{array}{c}P\left(\begin{array}{l}\text{firsttwoare}\\ \text{wrongsizes}\end{array}\right)=\frac{\text{Favorablacase}}{\text{Totalcases}}\\ =\frac{960}{6840}\\ =0.1404\end{array}$

(c)

To determine

To find: the probability that 1^st four shirts that pick are all extra-large.

Answer to Problem 18E

0.0031

Explanation of Solution

Given:

Total shirt = 20

Extra- large shirt = 6

Calculation:

The total number of favourable event is.

  $\begin{array}{c}\text{Total cases =}\left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 1}}^{st}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 2}}^{nd}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 3}}^{ed}\text{ time}\end{array}\right)\left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 4}}^{th}\text{ time}\end{array}\right)\\ =20\times 19\times 18\times 17\\ =116280\end{array}$

The favourable cases for first four selected shirts are extra-large is,

  $\begin{array}{c}\text{Favorablecases=}\left(\begin{array}{l}{1}^{st}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{2}^{ed}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{3}^{ed}\text{ shirtis }\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{4}^{th}\text{ shirtis }\\ \text{mediumsize}\end{array}\right)\\ =6\times 5\times 4\times 3\\ =360\end{array}$

Required probability is

  $\begin{array}{c}P\left(\begin{array}{l}\text{firsttwoare}\\ \text{wrongsizes}\end{array}\right)=\frac{\text{Favorablacase}}{\text{Totalcases}}\\ =\frac{360}{116,280}\\ =0.0031\\ \end{array}$

(d)

To determine

To find: the probability that at least one of the first four shirts check is a medium.

Answer to Problem 18E

0.6244

Explanation of Solution

Given:

Total number shirt = 20

Calculation:

  $\begin{array}{c}\text{Total cases =}\left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 1}}^{st}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 2}}^{nd}\text{ time}\end{array}\right)\times \left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 3}}^{ed}\text{ time}\end{array}\right)\left(\begin{array}{l}\text{Number of shirts}\\ {\text{when grabbing 4}}^{th}\text{ time}\end{array}\right)\\ =20\times 19\times 18\times 17\\ =116280\end{array}$

The favourable cases for less than 4 shirts are not medium is,

  $\begin{array}{c}\text{Favorablecases=}\left(\begin{array}{l}{1}^{st}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{2}^{ed}\text{shirtisnot}\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{3}^{ed}\text{ shirtis }\\ \text{mediumsize}\end{array}\right)\times \left(\begin{array}{l}{4}^{th}\text{ shirtis }\\ \text{mediumsize}\end{array}\right)\\ =16\times 15\times 14\times 13\\ =43680\end{array}$

Therefore the required probability is

  $\begin{array}{c}P\left(\begin{array}{l}\text{At least one of the }\\ \text{fourshirts is medium}\end{array}\right)=\text{1-}P\left(\begin{array}{l}\text{Less than 4 shirts}\\ \text{are medium}\end{array}\right)\\ =1-\left(\frac{43680}{116280}\right)\\ =1-03756\\ =0.6244\end{array}$