The probability of three independent events occurring are 1/2,1/3,1/4 respectively. What's the probability that only one of them will occur
Since A,B,C are independent, the chance that only A happens is P(A)*P(not B and not C) ...
So, P(only one event) is
1/2 * (2/3 * 3/4) + 1/3 * (1/2 + 3/4) + 1/4 * (1/2 * 2/3) = 3/4
Well, let's calculate the probability using my humorous approach! If you think about it, these events are like three friends trying to win a lottery.
First, let's imagine their chances as ticket numbers. The first friend has half the tickets, the second friend has a third of the tickets, and the third friend has a quarter of the tickets.
Now, to find the probability that only one of them will occur, we need to imagine an unlucky scenario where two friends don't win. Imagine the first friend's tickets being blown away, the second friend's tickets sinking in quicksand, and the third friend's tickets being eaten by a mischievous squirrel. Poor friends!
So, in this unlucky scenario, only one friend has tickets left, and that friend is the lucky winner! Since each friend has an equal chance of being the winner, we multiply the probabilities.
The probability that only one of them will occur is (1/2) * (2/3) * (3/4) = 1/4.
So, there you have it! The probability that only one of the events will occur is 1/4. But remember, in the world of probability, it's all about chances and surprises!
To find the probability that only one of the three independent events will occur, we need to consider the possibilities where only one event occurs while the other two do not.
Let's assume event A has a probability of 1/2, event B has a probability of 1/3, and event C has a probability of 1/4.
The probability that only event A occurs can be calculated by multiplying the probability of event A happening (1/2) with the probability of event B not happening (1 - 1/3) and the probability of event C not happening (1 - 1/4):
P(A only) = (1/2) * (1 - 1/3) * (1 - 1/4) = 1/2 * 2/3 * 3/4 = 1/4
Similarly, we can calculate the probabilities for only event B occurring and only event C occurring:
P(B only) = (1 - 1/2) * (1/3) * (1 - 1/4) = 1/2 * 1/3 * 3/4 = 1/8
P(C only) = (1 - 1/2) * (1 - 1/3) * (1/4) = 1/2 * 2/3 * 1/4 = 1/12
Finally, we add up these individual probabilities to find the probability that only one of the three events will occur:
P(only one event occurs) = P(A only) + P(B only) + P(C only) = 1/4 + 1/8 + 1/12 = 3/8
To find the probability that only one of the three events will occur, we can use the principle of inclusion-exclusion.
First, let's calculate the probability of each event occurring:
- Event A: P(A) = 1/2
- Event B: P(B) = 1/3
- Event C: P(C) = 1/4
To calculate the probability that only one event occurs, we need to consider the following possibilities:
1. Event A occurs, Event B & C do not occur
2. Event B occurs, Event A & C do not occur
3. Event C occurs, Event A & B do not occur
Let's calculate each possibility separately.
1. Event A occurs, Event B & C do not occur:
P(A) * (1 - P(B)) * (1 - P(C)) = (1/2) * (1 - 1/3) * (1 - 1/4)
2. Event B occurs, Event A & C do not occur:
P(B) * (1 - P(A)) * (1 - P(C)) = (1/3) * (1 - 1/2) * (1 - 1/4)
3. Event C occurs, Event A & B do not occur:
P(C) * (1 - P(A)) * (1 - P(B)) = (1/4) * (1 - 1/2) * (1 - 1/3)
Now, let's calculate each possibility:
1. (1/2) * (2/3) * (3/4)
2. (1/3) * (1/2) * (3/4)
3. (1/4) * (1/2) * (2/3)
Summing up the probabilities of the three cases, we get:
(1/2) * (2/3) * (3/4) + (1/3) * (1/2) * (3/4) + (1/4) * (1/2) * (2/3)
Simplifying the expression, we get:
6/24 + 3/24 + 4/24 = 13/24
Therefore, the probability that only one of the three events will occur is 13/24.