Question

at michaels school, 38% of the students have a pet dog and 24% of the students have a pet cat. michael found that 11% of the students had both a pet dog and a pet cat. what is the probability that a randomly - chosen student at michaels school will have a pet dog or a pet cat?
a. 83%
b. 40%
c. 51%
d. 62%

Explanation:

Step1: Recall probability formula

Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), where \(A\) is the event of having a pet - dog and \(B\) is the event of having a pet - cat.

Step2: Substitute given values

Let \(P(A) = 38\%\), \(P(B)=24\%\) and \(P(A\cap B)=11\%\). Then \(P(A\cup B)=38 + 24-11\).

Step3: Calculate result

\(P(A\cup B)=51\%\).

Answer:

C. 51%