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1) Two anti-aircraft guns are firing at 4 planes. P (they are firing at the same plane)?
2) Three anti-aircraft guns are firing at 4 planes. P (they are firing at the same plane)?
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Each anti air craft has 1/4 prob to be shooting at one plane, so 1/4*1/4 (two anti air craft). Then we multiply by 4 because we have 4 planes to choose
Each anti air craft has 1/4 prob to be shooting at one plane, so 1/4*1/4 (two anti air craft). Then we multiply by 4 because we have 4 planes to choose
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Mbao4,
Shouldn't the first one be multiplied by 6 instead of 4, since we have 4C2 ways to choose 2 planes out of 4?
Each anti air craft has 1/4 prob to be shooting at one plane, so 1/4*1/4 (two anti air craft). Then we multiply by 4 because we have 4 planes to choose
Mbao4,
Shouldn't the first one be multiplied by 6 instead of 4, since we have 4C2 ways to choose 2 planes out of 4?
Please explain
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Before Mba answers, let me try.
Think about two planes separately: the probability that one is firing is 100% or 1 (appears from the question stem). Now, the P that the second gun is firing at the same plane as the first one is 1/4, is't it? - four planes - one gun - this should be clear.
Multiply to get the total P. 1*1/4=1/4
Gabish?
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.