Two toys are drawn from a bag containing 21 toys numbered from 1 to 21 (without replacement). Find the probability that both the toys will show even numbers.

5/14

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9/14

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11/14

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3/14

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None of these

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Solution

The correct option is D
3/14

P(E) = $\frac{10 \times 9}{21 \times 20} = \frac{3}{14}$

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Two toys are drawn from a bag containing 21 toys numbered from 1 to 21 (without replacement). Find the probability that both the toys will show even numbers.

The correct option is D 3/14 P(E) = 10×921×20=314

The correct option is D
3/14

P(E) = $\frac{10 \times 9}{21 \times 20} = \frac{3}{14}$