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Each morning, coffee is brewed in the school workroom by one of three faculty members, depending on who arrives first at work. Mr. Worcester arrives first $10\%$ of the time, Dr. Currier arrives first $50\%$of the time, and Mr. Legacy arrives first on the remaining mornings. The probability that the coffee is strong when brewed by Dr. Currier is $0.1$, while the corresponding probabilities when it is brewed by Mr. Legacy and Mr. Worcester are $0.2$ and $0.3$, respectively. Mr. Worcester likes strong coffee!
(a) What is the probability that on a randomly selected morning the coffee will be strong?
(b) If the coffee is strong on a randomly selected morning, what is the probability that it was brewed by Dr. Currier?

Step by step solution

01

Part (a) Step 1: Given information

To see what the chances are that on a random morning, the coffee will be strong.

02

Explanation

It is assumed that one of the three faculty members brews coffee in the school workroom each morning, depending on who arrives first for work. Also,

$$\begin{gathered} P(Mr.Worcester)=0.10 \\ P(Dr.Currier)=0.50 \\ P(Strongcoffeeâˆ\pounds Dr.Currier)=0.1 \\ P(Strongcoffeeâˆ\pounds Mr.Worcester)=0.3 \\ P(Strongcoffeeâˆ\pounds Mr.Legacy)=0.2 \end{gathered}$$

Now we'll figure out what the chances are for Mr. Legacy, and that will be the remaining days besides the two. As a result,

$$\begin{gathered} P(Mr.Legacy)=1−0.1−0.5 \\ =0.4=40\% \end{gathered}$$

Using the multiplication rule, we can now calculate:

$$\begin{gathered} P(StrongcoffeeandDr.Currier) \\ =P(Dr.Currier)×P(Strongcoffeeâˆ\pounds Dr.Currier) \\ =0.50×0.1=0.05 \end{gathered}$$
$$\begin{gathered} P(StrongcoffeeandMr.Worcester) \\ =P(Mr.Worcester)×P(Strongcoffeeâˆ\pounds Mr.Worcester) \\ =0.1×0.3=0.03 \end{gathered}$$
localid="1654262381937" $$\begin{gathered} P(StrongcoffeeandMr.Legacy) \\ =P(Mr.Legacy)P(Strongcoffeeâˆ\pounds Mr.Legacy) \\ =0.4×0.2=0.08 \end{gathered}$$

Using the addition rule, we can now calculate the likelihood that the coffee will be strong on a randomly selected morning.
$$\begin{gathered} P(Strongcoffee) \\ =P(StrongcoffeeandDr.Currier)+P(StrongcoffeeandMr.Worcester) \\ +P(StrongcoffeeandMr.Legacy) \\ =0.05+0.03+0.08 \\ =0.16=16\% \end{gathered}$$

As a result, there's a $0.16$chance that the coffee on a random morning will be strong.

03

Part (b) Step 1: Given information

To find, if the coffee is served on a random morning, the possibility that it was prepared by Mr. Currier.

04

Explanation

It is assumed that one of the three faculty members brews coffee in the school workroom each morning, depending on who arrives first for work. Also,

$$\begin{gathered} P(Mr.Worcester)=0.10 \\ P(Dr.Currier)=0.50 \\ P(Strongcoffeeâˆ\pounds Dr.Currier)=0.1 \\ P(Strongcoffeeâˆ\pounds Mr.Worcester)=0.3 \\ P(Strongcoffeeâˆ\pounds Mr.Legacy)=0.2 \end{gathered}$$

Now we'll figure out what the chances are for Mr. Legacy, and that will be the remaining days besides the two. As a result,

$$\begin{gathered} P(Mr.Legacy) \\ =1−0.1−0.5 \\ =40\% \end{gathered}$$

Using the multiplication rule, we can now calculate:

$$\begin{gathered} P(Strongcoffee\&Dr.Currier) \\ =P(Dr.Currier)×P(Strongcoffeeâˆ\pounds Dr.Currier) \\ =0.50×0.1 \\ =0.05 \end{gathered}$$$$\begin{gathered} P(StrongcoffeeandMr.Worcester) \\ =P(Mr.Worcester)×P(Strongcoffeeâˆ\pounds Mr.Worcester) \\ =0.03 \end{gathered}$$

$$\begin{gathered} P(StrongcoffeeandMr.Legacy) \\ =P(Megacy)×P(Strongcoffeeâˆ\pounds Mr.Legacy) \\ =0.08 \end{gathered}$$

And there's a $0.16$chance that the coffee will be strong on a random morning.

localid="1654453666516" $$\begin{gathered} P(Dr.Currier|Strongcoffee) \\ =\frac{P(StrongcoffeeandDr.Currier)}{P(Strongcoffee)} \\ =0.3125 \\ =31.25\% \end{gathered}$$

If the coffee is strong on a randomly selected morning, the probability that it was brewed by Mr. Currier is $31.25$percent.

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