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The probabilities that three men hit a target are 1/6 , 1/4 , 1/3 respectively. Each shoots once the target what is the probability that exactly one of the them hits the target?
- a)23/96
- b)42/127
- c)19/48
- d)31/72
Correct answer is option 'D'. Can you explain this answer?
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The probabilities that three men hit a target are 1/6 , 1/4 , 1/3 resp...
To find the probability that exactly one of the three men hits the target, we can use the concept of independent events. The probability that an event occurs is equal to 1 minus the probability that it does not occur.
Let's calculate the probability that exactly one of the three men hits the target.
First, let's find the probability that the first man hits the target and the other two men do not hit the target. The probability that the first man hits the target is 1/6, and the probability that the other two men do not hit the target is (1 - 1/4) * (1 - 1/3) = 3/4 * 2/3 = 1/2. Therefore, the probability that exactly one of the three men hits the target with the first man hitting the target is (1/6) * (1/2) = 1/12.
Similarly, the probability that exactly one of the three men hits the target with the second man hitting the target is (1 - 1/6) * (1/4) * (1 - 1/3) = 5/6 * 1/4 * 2/3 = 5/36.
And the probability that exactly one of the three men hits the target with the third man hitting the target is (1 - 1/6) * (1 - 1/4) * (1/3) = 5/6 * 3/4 * 1/3 = 5/24.
Finally, we can add up these probabilities to get the total probability that exactly one of the three men hits the target:
(1/12) + (5/36) + (5/24) = 1/12 + 5/36 + 10/36 = (3 + 5 + 10) / 36 = 18/36 = 1/2.
Therefore, the probability that exactly one of the three men hits the target is 1/2.
The correct answer is not listed among the options provided.
Let's calculate the probability that exactly one of the three men hits the target.
First, let's find the probability that the first man hits the target and the other two men do not hit the target. The probability that the first man hits the target is 1/6, and the probability that the other two men do not hit the target is (1 - 1/4) * (1 - 1/3) = 3/4 * 2/3 = 1/2. Therefore, the probability that exactly one of the three men hits the target with the first man hitting the target is (1/6) * (1/2) = 1/12.
Similarly, the probability that exactly one of the three men hits the target with the second man hitting the target is (1 - 1/6) * (1/4) * (1 - 1/3) = 5/6 * 1/4 * 2/3 = 5/36.
And the probability that exactly one of the three men hits the target with the third man hitting the target is (1 - 1/6) * (1 - 1/4) * (1/3) = 5/6 * 3/4 * 1/3 = 5/24.
Finally, we can add up these probabilities to get the total probability that exactly one of the three men hits the target:
(1/12) + (5/36) + (5/24) = 1/12 + 5/36 + 10/36 = (3 + 5 + 10) / 36 = 18/36 = 1/2.
Therefore, the probability that exactly one of the three men hits the target is 1/2.
The correct answer is not listed among the options provided.
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The probabilities that three men hit a target are 1/6 , 1/4 , 1/3 respectively. Each shoots once the target what is the probability that exactly one of the them hits the target?a)23/96b)42/127c)19/48d)31/72Correct answer is option 'D'. Can you explain this answer?
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