# Conditional Probability Exercise: Different customer types dining at a restaurant.

An exercise in my textbook that I have simplified details:

An restaurant has $3$ types of customers: $A, B, C$. The frequency of customer type dining at the restaurant is $10\%, 40\%, 50\%$, respectively. $70\%$ of $C$ type order wine while dining. The proportions for $B$ and $A$ types are $50\%$ and $30\%$, respectively.

1. Find the probability that for any 2 random customers, both order wine.
2. Choose 2 random customers and both don't order wine. Find the probability that both are $C$ type customer.

My attempt:

1. Probability that a random customer orders wine:

$P = 0.3 \times 0,1 + 0,5 \times 0,4 + 0,7 \times 0,5 = 58\%$

Then the probability that for any 2 random customers, both order wine is:

$P = 0.58 \times 0.58 = 33.64\%$

1. First, the probability that at least 1 of 2 customers order wine is:

$P = 0.58 + 0.58 - 0.58 \times 0.58 = 82.36%$

Then the probability that both don't order wine is:

$P = 1 - 0.8236 = 17.64\%$

This is where I got stuck. I am confused with the detail of 2 random customers.

• More directly the probability that both don't order wine is $0.42\times0.42$. Advice: if there are percentages (I hate them) then immediately translate them. Eventually if you have the outcome then allow them to return. Commented Jun 7, 2018 at 7:11

## 1 Answer

$$P(\text{both of C}\mid\text{both don't order wine})P(\text{both don't order wine})=$$$$P(\text{both don't order wine}\mid\text{both of C})P(\text{both of C})$$

You already found $P(\text{both don't order wine})$.

Can you find $P(\text{both don't order wine}\mid\text{both of C})$ and $P(\text{both of C})$ as well?

• P(both don't order wine∣both of C) = $0.3^2$ = 0.09. P(both of C) = $0.5^2 = 0.25$. Is this correct? Commented Jun 7, 2018 at 7:24
• Yes, that is correct. Now substitute and you can find what you are after. Commented Jun 7, 2018 at 7:28