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# 1) Two anti-aircraft guns are firing at 4 planes. P (they

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Manager
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1) Two anti-aircraft guns are firing at 4 planes. P (they [#permalink]

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08 Aug 2003, 01:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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08 Aug 2003, 02:43
(1) 1*1/4=1/4

(2) 1*1/4*1/4=1/16

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Manager
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08 Aug 2003, 10:47
May I... explain? pls see previous message
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KL

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08 Aug 2003, 11:04
sure you may

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12 Aug 2003, 05:39
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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25 Aug 2003, 02:55
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?

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27 Aug 2003, 11:12
p_malh wrote:
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?

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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Re: Probablity Clarification-for Mba04   [#permalink] 27 Aug 2003, 11:12
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# 1) Two anti-aircraft guns are firing at 4 planes. P (they

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