Question

video lesson 5.4 example 3 - using a formula to find p(a and b)

1. out of 45 customers at a breakfast café, 42 customers bought either coffee or orange juice. there were 30 customers who bought orange juice and 40 customers who bought coffee. what is the probability that a randomly selected customer bought both coffee and orange juice?

Explanation:

Step1: Recall the principle of inclusion - exclusion for sets.

The formula for the number of elements in the union of two sets \(A\) and \(B\) is \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), where \(n(A\cup B)\) is the number of elements in \(A\) or \(B\), \(n(A)\) is the number of elements in \(A\), \(n(B)\) is the number of elements in \(B\), and \(n(A\cap B)\) is the number of elements in both \(A\) and \(B\). Let \(A\) be the set of customers who bought orange juice and \(B\) be the set of customers who bought coffee. We know that \(n(A\cup B) = 42\), \(n(A)=30\), \(n(B) = 40\). We need to find \(n(A\cap B)\) first.
Rearranging the formula: \(n(A\cap B)=n(A)+n(B)-n(A\cup B)\)
Substitute the values: \(n(A\cap B)=30 + 40-42=28\)

Step2: Calculate the probability.

The total number of customers is \(N = 45\). The probability \(P(A\cap B)\) is the number of customers who bought both divided by the total number of customers. So \(P(A\cap B)=\frac{n(A\cap B)}{N}\)
Substitute \(n(A\cap B) = 28\) and \(N = 45\): \(P(A\cap B)=\frac{28}{45}\approx0.622\) (or as a fraction \(\frac{28}{45}\))

Answer:

The probability is \(\frac{28}{45}\) (or approximately \(0.622\))