Question

in a class of 78 students, 32 have dogs, 25 have cats and 10 have both. what is the probability a randomly selected student has a cat or dog? p(cat or dog) = ?%

Explanation:

Step1: Use the principle of inclusion - exclusion

The formula for \( P(A\cup B) \) (where \( A \) is the event of having a dog and \( B \) is the event of having a cat) is \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \). First, we find the number of students who have a cat or a dog. The number of students with a dog is \( n(A) = 32 \), with a cat is \( n(B)=25 \), and with both is \( n(A\cap B) = 10 \). So the number of students with a cat or a dog is \( n(A\cup B)=32 + 25-10=47 \).

Step2: Calculate the probability

The total number of students is \( N = 78 \). So the probability \( P=\frac{47}{78}\approx0.6026 \). To convert to a percentage, we multiply by 100: \( 0.6026\times100 = 60.26\% \approx 60.3\% \) (or we can keep more decimal places as needed, but usually two decimal places for percentage in such cases).

Answer:

\( 60.3\% \) (or approximately \( 60.26\% \) if we use the exact fraction \( \frac{47}{78}\times100\approx60.256\% \))