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# Two apples and five bananas are defective out of 10 apples and 20 ban

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Two apples and five bananas are defective out of 10 apples and 20 ban [#permalink]
[Total ways to select 2 fruits = 30C2 ( this will be the base/denominator for all ) ]

P(both are good)
That is, it can be 2 good apples or 2 good bananas or one good apple and one good banana
There are 8 A and 15 B that are good, so selecting 2 out of these is 23C2/30C2
P=23∗22/30∗29=506/870

P(both are bananas)
That is, it can 2 good bananas, 2 bad bananas, one good, and one defective banana
There are 20 bananas totally, so selecting 2 out of these is 20C2
P=20*19/30*29=380/870

Notice that 2 good bananas are included in the calculations twice, hence you need to remove/subtract that probability from the total.

P(both are good bananas is repeated twice)
{Therefore we need to subtract 2 good bananas from the total so far which is 506/870 + 380/870 = 886/870}
There are 15 good bananas, so selecting 2 each out of this is 15C2/30C2
P=15*14/30*29 = 210/870 ( remember this has to be subtracted from the total )

Therefore the final answer is : P(both are good) + P(both are bananas) - P(both are good bananas is repeated twice)
=23C2/30C2 + 20C2/30C2 - 15C2/30C2
=506/870 + 380/870 - 210/870
=676/870 ( simplify by dividing by 2 )
=338/435
The answer is option B

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Re: Two apples and five bananas are defective out of 10 apples and 20 ban [#permalink]
Required : Both Bananas or both Good.

Both Bananas: 2 fruits drawn at random(Combinations) - 20C2(18 good, 2 defective)=190
Both Good: 2 apples can be good,2 Bananas can be good,1 Apple and 1 Banana Good.
2 Bananas Good already counted in 20c2, Hence 8c2+8c1*15c1=148

favorable outcomes 190+148=338/30c2
Re: Two apples and five bananas are defective out of 10 apples and 20 ban [#permalink]
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