Question

at a convention center, there are a total of 275 visitors. each visitor is located in either room a, room b or room c. if one of these visitors is selected at random, the probability of selecting a visitor who is located in room a is 0.68, and the probability of selecting a visitor who is located in room b is 0.24. how many visitors are located in room c? a. 8 b. 22 c. 45 d. 121

Explanation:

Step1: Use probability formula

Let $P(A) = 0.68$, $P(B)=0.24$. Since the sum of probabilities of all mutually - exclusive events (visitors are in room A, B or C) is 1, let the number of visitors in room C be $x$. The total number of visitors is $N = 275$. We know that $P(A)+P(B)+P(C)=1$. So $P(C)=1 - P(A)-P(B)$.
$P(C)=1 - 0.68 - 0.24=0.08$.

Step2: Calculate number of visitors in room C

We know that probability $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Let the number of visitors in room C be $n$. Then $P(C)=\frac{n}{N}$.
$n = P(C)\times N$. Substitute $P(C)=0.08$ and $N = 275$ into the formula.
$n=0.08\times275 = 22$.

Answer:

b. 22