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Konstantin Lynov
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1) Two anti-aircraft guns are firing at 4 planes. P (they 08 Aug 2003, 02:58
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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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1) Two anti-aircraft guns are firing at 4 planes. P (they 08 Aug 2003, 03:43
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(1) 1*1/4=1/4

(2) 1*1/4*1/4=1/16
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Konstantin Lynov
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May I... explain? pls see previous message
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sure you may
your explanations are welcome
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MBA04
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1) 1/4 * 1/4 *4 = 1/4
2) 1/4*1/4*1/4*4 = 1/16

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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1) Two anti-aircraft guns are firing at 4 planes. P (they 25 Aug 2003, 03:55
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Quote:
1) 1/4 * 1/4 *4 = 1/4
2) 1/4*1/4*1/4*4 = 1/16

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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p_malh
Quote:
1) 1/4 * 1/4 *4 = 1/4
2) 1/4*1/4*1/4*4 = 1/16

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


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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