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In a combat, A targets B, and both B and C target A. The probabilities of A, B, C hitting their targets are 2/3, 1/2 and 1/3 respectively. They shoot simultaneously and A is hit. Find the probability that B hits his target whereas C does not.
- a)1/2
- b)3/4
- c)2/5
- d)2/2
Correct answer is option 'A'. Can you explain this answer?
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In a combat, A targets B, and both B and C target A. The probabilities...
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In a combat, A targets B, and both B and C target A. The probabilities...
To find the probability that B hits his target whereas C does not, we need to consider the different scenarios in which A is hit.
There are three possible scenarios:
1. A hits B and C does not hit A.
2. A hits C and B does not hit A.
3. A hits both B and C.
Let's calculate the probability of each scenario:
Scenario 1: A hits B and C does not hit A
The probability of A hitting B is 2/3.
The probability of C not hitting A is 2/3 (since the complement of C hitting A is C not hitting A).
Therefore, the probability of scenario 1 is (2/3) * (2/3) = 4/9.
Scenario 2: A hits C and B does not hit A
The probability of A hitting C is 1/3.
The probability of B not hitting A is 1/2 (since the complement of B hitting A is B not hitting A).
Therefore, the probability of scenario 2 is (1/3) * (1/2) = 1/6.
Scenario 3: A hits both B and C
The probability of A hitting B is 2/3.
The probability of A hitting C is 1/3.
Therefore, the probability of scenario 3 is (2/3) * (1/3) = 2/9.
Now, let's calculate the probability that A is hit in total:
The probability of A being hit is the sum of the probabilities of all three scenarios:
(4/9) + (1/6) + (2/9) = 9/18 + 3/18 + 4/18 = 16/18 = 8/9.
Finally, to find the probability that B hits his target whereas C does not, we need to consider only scenario 1:
The probability of B hitting his target while C does not is the probability of scenario 1 happening, divided by the probability of A being hit:
(4/9) / (8/9) = 4/8 = 1/2.
Therefore, the correct answer is option 'A' (1/2).
There are three possible scenarios:
1. A hits B and C does not hit A.
2. A hits C and B does not hit A.
3. A hits both B and C.
Let's calculate the probability of each scenario:
Scenario 1: A hits B and C does not hit A
The probability of A hitting B is 2/3.
The probability of C not hitting A is 2/3 (since the complement of C hitting A is C not hitting A).
Therefore, the probability of scenario 1 is (2/3) * (2/3) = 4/9.
Scenario 2: A hits C and B does not hit A
The probability of A hitting C is 1/3.
The probability of B not hitting A is 1/2 (since the complement of B hitting A is B not hitting A).
Therefore, the probability of scenario 2 is (1/3) * (1/2) = 1/6.
Scenario 3: A hits both B and C
The probability of A hitting B is 2/3.
The probability of A hitting C is 1/3.
Therefore, the probability of scenario 3 is (2/3) * (1/3) = 2/9.
Now, let's calculate the probability that A is hit in total:
The probability of A being hit is the sum of the probabilities of all three scenarios:
(4/9) + (1/6) + (2/9) = 9/18 + 3/18 + 4/18 = 16/18 = 8/9.
Finally, to find the probability that B hits his target whereas C does not, we need to consider only scenario 1:
The probability of B hitting his target while C does not is the probability of scenario 1 happening, divided by the probability of A being hit:
(4/9) / (8/9) = 4/8 = 1/2.
Therefore, the correct answer is option 'A' (1/2).
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In a combat, A targets B, and both B and C target A. The probabilities of A, B, C hitting their targets are 2/3, 1/2 and 1/3 respectively. They shoot simultaneously and A is hit. Find the probability that B hits his target whereas C does not.a)1/2b)3/4c)2/5d)2/2Correct answer is option 'A'. Can you explain this answer?
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