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28 Jun 2021, 08:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
48% (03:20) correct
52%
(03:00)
wrong
based on 123
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Two apples and five bananas are defective out of 10 apples and 20 bananas contained in a fruit basket. If Sanjeev takes out two fruits at random, what is the probability that either both are bananas, or both are good?
(A) 119/435
(B) 338/435
(C) 841/870
(D) 217/870
(E) None of these
(A) 119/435
(B) 338/435
(C) 841/870
(D) 217/870
(E) None of these
28 Jun 2021, 20:23
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Bunuel
Total ways to select 2 fruits = 30C2
P(both are good)
There are 8 A and 15 B that are good, so selecting 2 out of these is 23C2
P=\(\frac{23*22}{30*29}=\frac{506}{870}\)
We already have ways to pick 2 good B in above, so we are left with the following two cases
P(both are defective bananas)
There are 5 B that are defective, so selecting 2 out of these is 5C2
P=\(\frac{5*4}{30*29}=\frac{20}{870}\)
P(one is defective B and other good B)
There are 5 defective B and 15 good B, so selecting 1 each out of these is 5C1*15C1
P=\(\frac{2*5*15}{30*29}=\frac{150}{870}\)
Total \(\frac{506+20+150}{870}=\frac{676}{870}=\frac{338}{435}\)
B
29 Jun 2021, 03:18
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We can Calculate P( Good_fruit) + P( Bananas) - P(Good+ Bananas) = (23C2 + 20C2 - 15C2 ) / 30C2 = 338/435
Bunuel
