Of 500 employees of a firm 280 have a life insurance policy ,400 have a medical insurance cover and 200 participate in both programs.

1)what is the probability that a randomly selected employee will be a participant in at least one of the two programs ?
2)Determine the probability that an employee will be a participant in the life insurance plan given that he/she has a medical insurance coverage?

Pr(one of them)=(280+400-200)/500

= 480/500

Pr(life given medical)=80/400

pr 480/500

1) Addition rule

(280/500)+(400/500)-(200/500)
=480/500

1) Well, if we want to find the probability that a randomly selected employee will be a participant in at least one of the two programs, we can add the probability of being in the life insurance program to the probability of being in the medical insurance program and then subtract the probability of being in both programs since we don't want to count them twice.

So, let's do some math here. We have 280 employees in the life insurance program, 400 employees in the medical insurance program, and 200 employees in both programs.

The probability of being in the life insurance program is 280/500 = 0.56.

The probability of being in the medical insurance program is 400/500 = 0.8.

Since 200 employees are in both programs, we subtract that from the total count: 500 - 200 = 300 employees.

So, the probability of being in at least one of the two programs is (0.56 + 0.8) - (200/300) = 1.36 - 0.67 = 0.69.

Therefore, the probability that a randomly selected employee will be a participant in at least one of the two programs is 0.69.

2) Now, let's move on to the second part of the question. We want to determine the probability that an employee will be a participant in the life insurance plan given that he/she has a medical insurance coverage.

Since we know that 400 employees have medical insurance coverage, and 200 employees are participants in both programs, we subtract the count of employees in both programs from the count of employees with medical insurance to get the count of employees with medical insurance only: 400 - 200 = 200 employees.

Out of those 200 employees, we know that 200 are also participants in the life insurance program. So, the probability of an employee having medical insurance coverage and participating in the life insurance program is 200/200 = 1.

Therefore, the probability that an employee will be a participant in the life insurance plan, given that he/she has medical insurance coverage, is 1.

To find the probability, we need to use the formula:

P(A or B) = P(A) + P(B) - P(A and B)

Let's calculate the probabilities based on the given information:

1) The probability that a randomly selected employee will be a participant in at least one of the two programs:

P(Life insurance) = 280/500 = 0.56
P(Medical insurance) = 400/500 = 0.8
P(Life insurance and Medical insurance) = 200/500 = 0.4

P(at least one program) = P(Life insurance) + P(Medical insurance) - P(Life insurance and Medical insurance)
= 0.56 + 0.8 - 0.4
= 0.96

Therefore, the probability that a randomly selected employee will be a participant in at least one of the two programs is 0.96 or 96%.

2) The probability that an employee will be a participant in the life insurance plan given that he/she has a medical insurance coverage:

P(Life insurance | Medical insurance) = P(Life insurance and Medical insurance) / P(Medical insurance)
= 200/400
= 0.5

Therefore, the probability that an employee will be a participant in the life insurance plan given that he/she has a medical insurance coverage is 0.5 or 50%.