Chapter 14, Problem 38E

a)

To determine

To find the probability that he gets stopped on Monday and again on Tuesday.

Answer to Problem 38E

The probability thathe gets stopped on Monday and again on Tuesday = 0.0225

Explanation of Solution

Given:

  $\begin{array}{l}P\left(X=x\right)=15\%=0.15\\ Where,\\ x=Monday,Tuesday,Wednesday,Thursday,Friday\end{array}$

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Using multiplication rule,

  $\begin{array}{l}P(MondayandTuesday)=0.15\times 0.15\\ P(MondayandTuesday)=0.0225\end{array}$

Hence, probability that he gets stopped on Monday and again on Tuesday is 0.0225

b)

To determine

To find the probability that he gets stopped first time on Thursday.

Answer to Problem 38E

The probability that he gets stopped first time on Thursday=0.0921

Explanation of Solution

Given:

  $\begin{array}{l}P\left(X=x\right)=15\%=0.15\\ Where,\\ x=Monday,Tuesday,Wednesday,Thursday,Friday\end{array}$

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Using complement rule,

  $\begin{array}{l}P(notwednesday)=P(notTuesday)=P(notMonday)=1-P(Friday)\\ P(notwednesday)=P(notTuesday)=P(notMonday)=1-0.15=0.85\end{array}$

Using multiplication rule,

  $\begin{array}{l}P(FirstonThursday)=0.85\times 0.85\times 0.85\times 0.15\\ P(FirstonThursday)=0.0921\end{array}$

Hence, probability that he gets stopped first time on Thursday is 0.0921

c)

To determine

To find the probability that he gets stopped every day.

Answer to Problem 38E

The probability that he gets stopped every day=0.000076

Explanation of Solution

Given:

  $\begin{array}{l}P\left(X=x\right)=15\%=0.15\\ Where,\\ x=Monday,Tuesday,Wednesday,Thursday,Friday\end{array}$

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Using multiplication rule,

  $\begin{array}{l}P(everyday)=P(Monday)\times P(Tuesday)\times P(wednesday)\times P(Thursday)\times P(Friday)\\ P(everyday)=0.15\times 0.15\times 0.15\times 0.15\times 0.15\\ P(everyday)=0.000077\end{array}$

Hence, probability that he gets stopped every day is 0.000076

d)

To determine

To find the probability that he gets stopped at least once during the week.

Answer to Problem 38E

The probability that he gets stopped at least once during the week = 0.5563

Explanation of Solution

Given:

  $\begin{array}{l}P\left(X=x\right)=15\%=0.15\\ Where,\\ x=Monday,Tuesday,Wednesday,Thursday,Friday\end{array}$

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Using multiplication rule,

  $\begin{array}{l}P(never)=P(NotMonday)\times P(NotTuesday)\times P(Notwednesday)\times P(NotThursday)\times P(NotFriday)\\ P(never)=0.85\times 0.85\times 0.85\times 0.85\times 0.85\\ P(never)=0.4437\end{array}$

Using complement rule,

  $\begin{array}{l}P(atleastonce)=1-P(never)\\ P(atleastonce)=1-0.4437=0.5563\end{array}$

Hence, probability that he gets stopped at least once during the week is 0.5563