Question

a survey of 200 people found that 112 people own a phone, 76 own a laptop, and 50 own both. what is the probability that a randomly selected person owns a laptop or a phone? p(laptop or phone) = ?%

Explanation:

Step1: Recall the formula for the probability of the union of two events

The formula for \( P(A \cup B) \) (the probability that \( A \) or \( B \) occurs) is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \). Here, let \( A \) be the event that a person owns a phone and \( B \) be the event that a person owns a laptop. We know the total number of people \( n = 200 \), \( n(A)=112 \), \( n(B) = 76 \), and \( n(A\cap B)=50 \). First, we find the probabilities \( P(A)=\frac{n(A)}{n}\), \( P(B)=\frac{n(B)}{n}\), and \( P(A\cap B)=\frac{n(A\cap B)}{n} \).

Step2: Calculate the number of people who own a laptop or a phone

Using the principle of inclusion - exclusion for the number of elements in the union, the number of people who own a laptop or a phone is \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \). Substituting the given values: \( n(A\cup B)=112 + 76-50=138 \).

Step3: Calculate the probability

The probability \( P(\text{Laptop or Phone})=\frac{n(A\cup B)}{n}\times100\% \) (to get the percentage). Substituting \( n(A\cup B) = 138 \) and \( n = 200 \), we have \( P(\text{Laptop or Phone})=\frac{138}{200}\times100\%=69\% \).

Answer:

\( 69\% \)