Question

Let x represent the number of times a student visits a gym in a one month period. Assume that the probability distribution of X is as follows: x 0 1 2 3 p(x) 0.17 0.33 0.32 0.18 Determine the probability the student visits the gym at most twice in a month. Report your answer to two decimal places.

Answer 1

Answer: Probability of visiting at most twice = 0.82

Explanation: The probability distribution is of the form:

X 0 1 2 3

P(X) 0.17 0.33 0.32 0.18

It wants the probability of visiting the gym at most twice in a month, which means the probability of never going to the gym, P(X=0), or going once, P(X=1), or going twice, P(X=2).

Using the "OR" probability:

P(visiting at most twice) = P(X=0) + P(X=1) + P(X=2)

P(visiting at most twice) = 0.17 + 0.33 + 0.32

P(visiting at most twice) = 0.82

Therefore, the probability of visiting the gym at most twice in a month is 0.82 or 82%