Question

select the correct answer. 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. 63% b. 40% c. 51% d. 62%

Explanation:

Step1: Recall the probability formula for union

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 the given values

Let $P(A) = 38\%=0.38$, $P(B)=24\% = 0.24$ and $P(A\cap B)=11\%=0.11$. Then $P(A\cup B)=0.38 + 0.24-0.11$.

Step3: Calculate the result

$P(A\cup B)=0.38+0.24 - 0.11=0.51 = 51\%$.

Answer:

C. 51%