RAMU can hit a target 3 times in 6 shots, SIVA can hit the target 2 times in 6 shots and SHYAM can hit 4 times in 4 shots. What is the probability that at least 2 shots hit the targetOptions5/62/35/21/3
Question
RAMU can hit a target 3 times in 6 shots, SIVA can hit the target 2 times in 6 shots and SHYAM can hit 4 times in 4 shots. What is the probability that at least 2 shots hit the targetOptions5/62/35/21/3
Solution
The problem is asking for the probability that at least 2 shots hit the target. This means we need to find the probability that either 2, 3, or all 4 shots hit the target.
First, let's calculate the individual probabilities for each person:
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RAMU: He can hit the target 3 times in 6 shots. So, the probability of RAMU hitting the target is 3/6 = 0.5.
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SIVA: He can hit the target 2 times in 6 shots. So, the probability of SIVA hitting the target is 2/6 = 0.33.
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SHYAM: He can hit the target 4 times in 4 shots. So, the probability of SHYAM hitting the target is 4/4 = 1.
Next, we need to find the probability that at least 2 shots hit the target. This can happen in several ways:
- RAMU and SIVA hit the target
- RAMU and SHYAM hit the target
- SIVA and SHYAM hit the target
- All three hit the target
We can calculate these probabilities as follows:
- RAMU and SIVA: 0.5 * 0.33 = 0.165
- RAMU and SHYAM: 0.5 * 1 = 0.5
- SIVA and SHYAM: 0.33 * 1 = 0.33
- All three: 0.5 * 0.33 * 1 = 0.165
Adding these probabilities together gives us the total probability that at least 2 shots hit the target:
0.165 + 0.5 + 0.33 + 0.165 = 1.16
However, this number is greater than 1, which is not possible for a probability. This is because we have double-counted some scenarios where more than two people hit the target.
The correct way to solve this problem is to use the principle of inclusion and exclusion:
- First, add the probabilities for each pair of people: 0.165 (RAMU and SIVA) + 0.5 (RAMU and SHYAM) + 0.33 (SIVA and SHYAM) = 0.995
- Then, subtract the probability that all three people hit the target: 0.995 - 0.165 = 0.83
So, the probability that at least 2 shots hit the target is 0.83, or approximately 5/6. This is not one of the options given in the problem, so there may be a mistake in the problem or in my calculations.
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