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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.

Re: Probabilty Help [#permalink]
07 Apr 2011, 11:23

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

Re: Probabilty Help [#permalink]
07 Apr 2011, 11:45

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.

Re: Probabilty Help [#permalink]
07 Apr 2011, 11:57

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.

Re: Probabilty Help [#permalink]
08 Apr 2011, 08:00

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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