What is the probability that he or she slept for at least 8 hours last night? (Please round your final answer to 2 decimal places)

Question

Employees from a company accepted an interview talking about their sleep last night. It is found that 80% of the employees slept for at least 8 hours last night. Among those sleeping for at least 8 hours, 70% went to bed at 11 pm, while among those sleeping for less than 8 hours, only 30% went to bed at 11 pm. Suppose a randomly chosen employee is found to go to bed at 11 pm. What is the probability that he or she slept for at least 8 hours last night? (Please round your final answer to 2 decimal places)

Answer 1

Lets define our events first:

A is the event that the employee slept for at least 8 hours

E is the event that the employee went to bed at 11 PM

We are asked to find P(A | E), or the probability that an employee slept for at least 8 hours given that they went to bed at 11 pm. Lets start with the conditional probability formula and then expand from there:

P(A | E)=P(A and E) / P(E)

The problem now is that we don't know P(A and E) or P(E), but we can use the information we do have to figure it out! Lets start with P(A and E). This is the probability that an employee slept for 8 hours and went to bed at 11. Since we are given P(A)=80% and P(E | A)=70%, we can re-arrange our conditional probability formula to get

P(A and E) = P(E | A) * P(A) = .7*.8 = 0.56 (56%)

Now lets move to P(E). We can use the law of total probability to calculate this! Let A^c be the complement of A (the event that the employee gets less than 8 hours of sleep).

P(E) = P(A and E) + P(A^c and E) = 0.56 + P(E | A^c) * P(A^c) = 0.56 + 0.3 * (1-0.8) = 0.56 + 0.06 = 0.62 (62%)

Therefore, P(A|E) = 0.56 / 0.62 = 0.9032 or 90.32%