A box contains 6 bottles of variety1 drink 3 bottles of variety2 drink and 4 bottles of variety3 dri...........

Quantitative Aptitude ( CCS) Probability

A box contains 6 bottles of variety1 drink, 3 bottles of variety2 drink and 4 bottles of variety3 drink. Three bottles of them are drawn at random, what is the probability that the three are not of the same variety?

833/858

752/833

632/713

none of these

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Explanation:

Total number of drink bottles = 6 + 3 + 4 = 13.

Let S be the sample space.
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3 = (13 x 12 x 11)/(1 x 2 x 3) = 66 x 13 = 858.

Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= 6 x 5 x 4 / 1 x 2 x 3 + 1 + 4 x 3 x 2 / 1 x 2 x 3
= 20 + 1 + 4 = 25.

The probability of taking 3 bottles of the same variety = n(E)/n(S) = 25/858.

Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858 = 833/858.
Hence, option 1 is the correct answer.

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