atleast 4 head = 4H 1T + 5 Hbillyr wrote:Need help solving this prob question.
thanks
= 5C4 * (0.6)^4 (0.4) + 5C5*(0.6)^5
= 5 (0.6)^4 (0.4) + 0.6)^5
E
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Probability
Source: Beat The GMAT — Problem Solving |
= 5C4 * (0.6)^4 (0.4) + 5C5*(0.6)^5
= 5 (0.6)^4 (0.4) + 0.6)^5
E
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
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
