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18 Apr 2022, 09:22
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A basket contains 3 apples and 3 mangoes. If you select 3 pieces of fruit at random and without replacement, what is the probability that you’ll select at least one mango?
(A) \(\frac{1}{8}\)
(B) \(\frac{1}{10} \)
(C) \(\frac{3}{20}\)
(D) \(\frac{7}{8} \)
(E) \(\frac{19}{20}\)
(A) \(\frac{1}{8}\)
(B) \(\frac{1}{10} \)
(C) \(\frac{3}{20}\)
(D) \(\frac{7}{8} \)
(E) \(\frac{19}{20}\)
18 Apr 2022, 10:11
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total ways to select 3 pieces from 6 ; 6c3 ; 20
atleast 1 mango
3c1*3c2 + 3c2*3c1 + 3c3 = 19
19/20
option E
atleast 1 mango
3c1*3c2 + 3c2*3c1 + 3c3 = 19
19/20
option E
GMATNinja
A basket contains 3 apples and 3 mangoes. If you select 3 pieces of fruit at random and without replacement, what is the probability that you’ll select at least one mango?
(A) \(\frac{1}{8}\)
(B) \(\frac{1}{10} \)
(C) \(\frac{3}{20}\)
(D) \(\frac{7}{8} \)
(E) \(\frac{19}{20}\)
(A) \(\frac{1}{8}\)
(B) \(\frac{1}{10} \)
(C) \(\frac{3}{20}\)
(D) \(\frac{7}{8} \)
(E) \(\frac{19}{20}\)
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18 Apr 2022, 22:11
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We need to find the probability of choosing atleast one mango out of three choices that we made.
We can either calculate all the cases where one mango has been chosen or subtract the probability of choosing 'no' mango in three attempts and then subtract that from \(1\).
The latter method is better and quicker.
Thus, the probability of not getting even one mango in three attempts is equal to getting all apples in \(3 \) attempts.
Probability of getting \(3 \) apples = \(\frac{3}{6}* \frac{2}{5} * \frac{1}{4} = \frac{1}{20}\).
Thus, Probability of getting atleast one mango = \(1 - \frac{1}{20} = \frac{19}{20}\).
We can either calculate all the cases where one mango has been chosen or subtract the probability of choosing 'no' mango in three attempts and then subtract that from \(1\).
The latter method is better and quicker.
Thus, the probability of not getting even one mango in three attempts is equal to getting all apples in \(3 \) attempts.
Probability of getting \(3 \) apples = \(\frac{3}{6}* \frac{2}{5} * \frac{1}{4} = \frac{1}{20}\).
Thus, Probability of getting atleast one mango = \(1 - \frac{1}{20} = \frac{19}{20}\).
