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Probabilty Help

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anuu
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Probabilty Help [#permalink] New post   07 Apr 2011, 10:45
Suppose 5 % of Channel inhabitants are Cricket fans. Determine the approximate probability that a sample of 100 inhabitants will contain at least 3 cricket fans.

Can somebody help me out with the logic..

The problem states out of 100 people , atleast 3 should be cricket fans,

there r 2 posiibilities, 'cricket fans' and 'not cricket fans'

P(cricket fans) = 5/100 = 1/20

P(not cricket fans) = 95/100 = 19/20

Total number of possibilities = 2 ^100

I'm stuck here...

Can anybody pls help me with the logic? I came across this problem in 'urch' website.

Anu



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Re: Probabilty Help [#permalink] New post   07 Apr 2011, 11:12
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5% of 100 = 5 are cricket fans.

Three cases - 3 fans, 4 fans and 5 fans

Numbers of ways to choose 3 fans = 5C3
Numbers of ways to choose 4 fans = 5C4
Numbers of ways to choose 5 fans = 5C5

Hence probability = (5C3 + 5C4 + 5C5) / 100 = (10 + 5 + 1) / 100 = 16 / 100 = 0.16
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Re: Probabilty Help [#permalink] New post   07 Apr 2011, 11:46
Thanks!! for the quick reply. That turned out to be an easy appraoch.

My understanding was out of 100 any 5 could be cricket fans and the remaining 95 'not cricket fans'.

Suppose cricket fans are F and not cricket fans are N then

Case1 : FFFNNNN..(F 3times & N 97 times) ,,then there wud be many such cases for this kind of arrangement.. 100!/97!3! ways to arrange...

1/20*1/20*1/20*19/20...97times

case 2: FFFFNNN..(F 4times and & N 96times )..100!/4!96! arrangements...

case3 : FFFFFNNN(F 5 times and N 95 times)...

I was following this approach and i got stuck for some reason...i guess too many multipications to deal with:-)

But your approach made this problem simpler...Thanks!!

Anuu
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Re: Probabilty Help [#permalink] New post   07 Apr 2011, 12:23
I think there are 100C5 groups. So essentially the numerator must be divided by 100C5 and not just 100. So the probability becomes (5C3 + 5C4 + 5C5) / 100C5
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Re: Probabilty Help [#permalink] New post   07 Apr 2011, 12:45
As per my understanding , the denominator should be 100.

P(atlaest 3 cricket fans) = Number of possible outcome/Total number of outcomes

the problems states " determine the approximate probability that a sample of 100 inhabitants will contain at least 3 cricket fans.

if the denominator is 100c5, it implies that the number (number of ways) of channel inhabitants who r cricket fans ar 100c5. or the number of ways cricket fans can be arranged is 100c5.

Pls correct me if i'm wrong?

Anuu
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Re: Probabilty Help [#permalink] New post   07 Apr 2011, 12:57
I think a pool of 5 fans can be chosen in 100C5 ways. And you can sub-select within the 5 fans - 3 cases. i.e. 3 fans + 4 fans + 5 fans i.e. 5C3 * 95C2 + 5C4* 95C1 + 5C5 * 95C0. I think fluke can confirm this.
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Re: Probabilty Help [#permalink] New post   08 Apr 2011, 09:00
Anuu, 100C5 necessarily implies that the number (number of ways) of channel inhabitants who r cricket fans as its mentioned 5% only out of 100.



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Re: Probabilty Help [#permalink]
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