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nitinkarnwal
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07 Apr 2018, 13:56
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A basket contains 10 fruits of which 4 are rotten. If 3 fruits are drawn from the basket, find the probability that none of the fruits drawn is rotten.
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07 Apr 2018, 14:51
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Out of 10 fruits 6 are non rotten.
For fruits to be drawn 3 times,
Out of 6 fruits is same as
Drawing out of 6 fruits first time
So Prob= 6/10 = 3/5
Drawing out of remaining 5 fruits second time
So prob = 5/9
Drawing third time out of remaining 8fruits
So Prob = 4/8=1/2
To get Total Probability we multiply
3/5 x 5/9 x 1/2 = 1/6
Sent from my iPhone using GMAT Club Forum
For fruits to be drawn 3 times,
Out of 6 fruits is same as
Drawing out of 6 fruits first time
So Prob= 6/10 = 3/5
Drawing out of remaining 5 fruits second time
So prob = 5/9
Drawing third time out of remaining 8fruits
So Prob = 4/8=1/2
To get Total Probability we multiply
3/5 x 5/9 x 1/2 = 1/6
Sent from my iPhone using GMAT Club Forum
nitinkarnwal
Joined: 07 Sep 2016
Last visit: 08 Jun 2024
Posts: 8
Own Kudos:
Given Kudos: 119
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08 Apr 2018, 00:52
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What I think is:
According to the language of the question, 3 fruits are drawn simultaneously from the basket and are not drawn one by one.
Hence, P(none of the fruits drawn is rotten) = P (drawing a non-rotten fruit)= 6C3/10C3 = 1/6
According to the language of the question, 3 fruits are drawn simultaneously from the basket and are not drawn one by one.
Hence, P(none of the fruits drawn is rotten) = P (drawing a non-rotten fruit)= 6C3/10C3 = 1/6
