of 500 employees a firm 280 have a life insurance policy, 400 have a medical insurance cover and 200 participate in both programmes. what is the probability that a randomly selected employee will be a participant in at least one of the two programs
(ii) determine the probability that an employee will be a participant in the life insurance plan given that he/she has a medical insurance coverage.
(i) To find the probability that a randomly selected employee will be a participant in at least one of the two programs, we can use the principle of inclusion-exclusion.
Let's calculate the number of employees who are participants in at least one of the two programs:
Number of participants in life insurance = 280
Number of participants in medical insurance = 400
Number of participants in both programs = 200
To find the number of employees who participate in at least one of the two programs, we can add the number of participants in each program and then subtract the number of participants in both programs to avoid double counting:
Number of participants in at least one of the two programs = (Number of participants in life insurance) + (Number of participants in medical insurance) - (Number of participants in both programs)
Number of participants in at least one of the two programs = 280 + 400 - 200
Number of participants in at least one of the two programs = 480
Now, we can calculate the probability:
Probability = Number of participants in at least one of the two programs / Total number of employees
Probability = 480 / 500
Probability ≈ 0.96
Therefore, the probability that a randomly selected employee will be a participant in at least one of the two programs is approximately 0.96.
(ii) To find the probability that an employee will be a participant in the life insurance plan given that he/she has medical insurance coverage, we can use conditional probability.
Let's calculate the probability:
Probability = Number of participants in both programs / Number of participants in medical insurance
Probability = 200 / 400
Probability = 0.5
Therefore, the probability that an employee will be a participant in the life insurance plan given that he/she has medical insurance coverage is 0.5.
To find the probability that a randomly selected employee will be a participant in at least one of the two programs, we can use the principle of inclusion-exclusion.
Let's define:
A: Event that an employee has a life insurance policy.
B: Event that an employee has a medical insurance cover.
We are given:
P(A) = 280/500 = 0.56 (probability of having a life insurance policy)
P(B) = 400/500 = 0.8 (probability of having a medical insurance cover)
P(A ∩ B) = 200/500 = 0.4 (probability of participating in both programs)
To find the probability of participating in at least one of the two programs (P(A ∪ B)), we can use the formula:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 0.56 + 0.8 - 0.4
P(A ∪ B) = 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.
(ii) 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 (P(A|B)), we can use the formula:
P(A|B) = P(A ∩ B) / P(B)
P(A|B) = 0.4 / 0.8
P(A|B) = 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.
To find the probability that a randomly selected employee will be a participant in at least one of the two programs, we need to add the probabilities of two separate events: being a participant in the life insurance program and being a participant in the medical insurance program, then subtract the probability of being a participant in both programs.
Let's break down the information given:
- Number of employees in the firm (Total) = 500
- Number of employees with a life insurance policy (A) = 280
- Number of employees with medical insurance cover (B) = 400
- Number of employees participating in both programs (A ∩ B) = 200
(i) Probability of being a participant in at least one of the two programs:
To find this probability, we need to use the principle of inclusion-exclusion. It states that the probability of the union of two events (A ∪ B) is the sum of their individual probabilities minus the probability of their intersection (A ∩ B).
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = (Number of employees with a life insurance policy / Total employees) + (Number of employees with medical insurance cover / Total employees) - (Number of employees participating in both programs / Total employees)
P(A ∪ B) = (280/500) + (400/500) - (200/500)
P(A ∪ B) = 0.56 + 0.8 - 0.4
P(A ∪ B) = 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%.
(ii) Probability of being a participant in the life insurance plan given that he/she has medical insurance coverage (P(A|B)):
To find this probability, we can use the formula for conditional probability:
P(A|B) = P(A ∩ B) / P(B)
Given:
- Number of employees with a life insurance policy (A) = 280
- Number of employees with medical insurance cover (B) = 400
- Number of employees participating in both programs (A ∩ B) = 200
P(A|B) = (Number of employees participating in both programs / Number of employees with medical insurance cover)
P(A|B) = (200/400)
P(A|B) = 0.5
Therefore, the probability that an employee will be a participant in the life insurance plan, given that he/she has medical insurance coverage, is 0.5 or 50%.