5 apples and 6 oranges are kept in a box. If three fruits are chosen at random, then the probability that 2 apples and one orange are picked is:

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  1. \(\frac{4}{11}\)
  2. \(\frac{4}{13}\)
  3. \(\frac{5}{11}\)
  4. \(\frac{6}{11}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{4}{11}\)
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Detailed Solution

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Given:

5 apples and 6 oranges are kept in a box

Concept:

The no of way for choosing r objects from n no of objects is nCr .

Calculation:

5 apples and 6 oranges are kept in a box.

The no of way for choosing 2 apples from 5 apples is 5C2 = 10

The no of way for choosing 1 orange from 6 oranges is 6C= 6

The total no of way for choosing 3 fruits from 11 fruits is 11C3 = 165

Now,

The  probability that 2 apples and one orange are picked is \(\rm \frac{10\times6}{165}=\frac{4}{11}\)

Hence option (1) is correct.

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