Seven balls are drawn simultaneously from a bag containing 5 white balls are 6 green balls. The probability of drawing 3 white and 4 green balls is
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a
7 11C7
b
5C3+ 6C4 11C7
c
5C2× 6C2 11C7
d
6C3+ 5C4 11C7
answer is C.
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Detailed Solution
The number of ways selecting 7 balls from 11 balls is C(11,7) Suppose that E be the event of getting 3 white and 4 green balls n(E) =C(5,3)·C(6,4) =C(5,2)·C(6,2) Therefore, the probability of the event E is P(E) =n(E)n(S)=C(5,2)C(6,2)C(11,7)