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Probability

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Re: Probability

by x2suresh » Thu Feb 26, 2009 10:26 am
billyr wrote:Need help solving this prob question.


thanks
atleast 4 head = 4H 1T + 5 H
= 5C4 * (0.6)^4 (0.4) + 5C5*(0.6)^5

= 5 (0.6)^4 (0.4) + 0.6)^5


E
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by krisraam » Thu Feb 26, 2009 1:01 pm
Answer is E

Atleast 4Heads mean ...

we have these two possibilities = 4H 1T or 5H

The events are mutually exclusive.

probability of getting 5H = (0.6)^5.

probability of getting 4H1T =(5!/4!) [(0.6)^4* (0.4)]

Like HHHHT, THHHH etc. basically we are arranging 4 heads and 1 tail not selecting 4 heads out of 5. In that we are selecting 1 tail out of 5...

Lemme know if was wrong.

Thanks
raama
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