The probability of three independent events are 1/2,1/3,1/4 respectively. What's the probability that two of them will occur?
since the combinations are mutually exclusive, the final probability is the sum of the probabilities of the three combinations. That is,
1/2 * 1/3 + 1/2 * 1/4 + 1/3 * 1/4
To find the probability that exactly two out of three independent events will occur, we need to consider all possible combinations.
Let's call the events A, B, and C, with probabilities of 1/2, 1/3, and 1/4, respectively.
To calculate the probability that exactly two of the events occur, we need to consider the following scenarios:
1. A and B occur, but C does not.
2. A and C occur, but B does not.
3. B and C occur, but A does not.
The probability of scenario 1 happening is calculated as follows:
P(A and B happening) = P(A) * P(B) = (1/2) * (1/3) = 1/6
The probability of scenario 2 happening is calculated as follows:
P(A and C happening) = P(A) * P(C) = (1/2) * (1/4) = 1/8
The probability of scenario 3 happening is calculated as follows:
P(B and C happening) = P(B) * P(C) = (1/3) * (1/4) = 1/12
Since these events are mutually exclusive (i.e., they cannot occur at the same time), we can add up the probabilities of each scenario to find the total probability:
Total probability = P(A and B happening) + P(A and C happening) + P(B and C happening)
= (1/6) + (1/8) + (1/12)
= 4/24 + 3/24 + 2/24
= 9/24
= 3/8
Therefore, the probability that exactly two out of the three events will occur is 3/8.