91Ó°ÊÓ

We need to find the probability that on a randomly selected morning the coffee will be strong

Multiplication rule:

$P\left(AandB\right)=P\left(Aâˆ\copyright B\right)=P\left(A\right)×P\left(Bâˆ\pounds A\right)=P\left(B\right)×P\left(Aâˆ\pounds B\right)$

Addition rule for disjoint or mutually exclusive events:

$P\left(A∪B\right)=P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

$P\left(Mr.Worcester\right)=10\text{\%}=0.10$

localid="1650619178904" $P\left(Dr.Currier\right)=50\text{\%}=0.50$

$$\begin{gathered} P\left(StrongcoffeeDr.Currier\right)=0.1 \\ P\left(Strongcoffeeâˆ\pounds Mr.Worcester\right)=0.3 \\ P(Strongcoffeeâˆ\pounds Mr.Legacy)=0.2 \end{gathered}$$

Mr. Worcester arrives first $10\%$ of the time and Dr. Currier arrives first $50\text{\%}$,so the time, while Mr. Legacy arrives first on the remaining days and this is then $100\%-50\%-10\%=40\%$of the time:

localid="1650619365441" $P\left(\text{Mr}.Legacy\right)=40\text{\%}=0.40$

Use the general multiplication rule is :

localid="1650619862930" $$\begin{gathered} P\left(StrongcoffeeandDr.Currier\right)=P\left(Dr.Currier\right)×P\left(Strongcoffeeâˆ\pounds Dr.Currier\right) \\ =0.50×0.1 \\ =0.05 \\ P\left(StrongcoffeeandMr.Worcester\right)=P\left(Mr.Worcester\right)×P(Strongcoffeeâˆ\pounds Mr.Worce) \\ =0.10×0.3 \\ =0.03 \\ P(strongcoffeeandMr.Legacy)=P\left(Mr.Legacy\right)×P\left(strongcoffeeâˆ\pounds Mr.Legacy\right) \\ =0.40×0.2 \\ =0.08 \end{gathered}$$

Now, the addition rule of disjoint events:

$$\begin{gathered} P\left(Strongcoffee\right)=P\left(StrongcoffeeandDr.Currier\right)×P\left(StrongcoffeeandMr.Worcester\right)+P\left(StrongcoffeeandMr.\text{Legacy}\right) \\ =0.5+0.3+0.8 \\ =0.16 \\ =16\% \end{gathered}$$

We need to find the probability that it was brewed by Dr. Currier.

Multiplication rule:

$P\left(AandB\right)=P\left(Aâˆ\copyright B\right)=P\left(A\right)×P\left(Bâˆ\pounds A\right)=P\left(B\right)×P\left(Aâˆ\pounds B\right)$

Addition rule for disjoint or mutually exclusive events:

$P\left(A∪B\right)=P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

$P\left(Mr.Worcester\right)=10\text{\%}=0.10$

$P\left(Dr.Currier\right)=50\text{\%}=0.50$

$$\begin{gathered} P\left(StrongcoffeeDr.Currier\right)=0.1 \\ P\left(Strongcoffeeâˆ\pounds Mr.Worcester\right)=0.3 \\ P(Strongcoffeeâˆ\pounds Mr.Legacy)=0.2 \end{gathered}$$

Mr. Worcester arrives first $10\%$ of the time and Dr. Currier arrives first $50\%$,so the time, while Mr. Legacy arrives first on the remaining days and this is then $100\%-50\%-10\%=40\%$of the time:

$P\left(\text{Mr}.Legacy\right)=40\text{\%}=0.40$

Use the general multiplication rule is :

$$\begin{gathered} P\left(StrongcoffeeandDr.Currier\right)=P\left(Dr.Currier\right)×P\left(Strongcoffeeâˆ\pounds Dr.Currier\right) \\ =0.50×0.1 \\ =0.05 \\ P\left(StrongcoffeeandMr.Worcester\right)=P\left(Mr.Worcester\right)×P(Strongcoffeeâˆ\pounds Mr.Worce) \\ =0.10×0.3 \\ =0.03 \\ P(strongcoffeeandMr.Legacy)=P\left(Mr.Legacy\right)×P\left(strongcoffeeâˆ\pounds Mr.Legacy\right) \\ =0.40×0.2 \\ =0.08 \end{gathered}$$

Now, the addition rule of disjoint events:

$$\begin{gathered} P\left(Strongcoffee\right)=P\left(StrongcoffeeandDr.Currier\right)×P\left(StrongcoffeeandMr.Worcester\right)+P\left(StrongcoffeeandMr.\text{Legacy}\right) \\ =0.5+0.3+0.8 \\ =0.16 \\ =16\% \end{gathered}$$

Use the definition of conditional probability:

$$\begin{gathered} P(Dr.Currier|Strongcoffee)=\frac{P\left(StrongcoffeeandDr.Currier\right)}{P\left(Strongcoffee\right)} \\ =\frac{0.05}{0.16} \\ =\frac{5}{16} \\ =0.3125 \\ =31.25\% \end{gathered}$$