A bag contains 21 toys numbered 1 to 21. A to | vedclass

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(B) Total number of toys is $21$. The even numbers between $1$ and $21$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$. There are $10$ even numbers.Probabil

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$\frac{3}{14}$

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(B) Total number of toys is $21$. The even numbers between $1$ and $21$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$. There are $10$ even numbers.Probability of drawing the first even number is $\frac{10}{21}$.After drawing one even number,there are $9$ even numbers left out of $20$ total toys.Probability of drawing the second even number is $\frac{9}{20}$.The probability that both toys show even numbers is $\frac{10}{21} 	imes \frac{9}{20} = \frac{90}{420} = \frac{3}{14}$.

$A$ bag contains $21$ toys numbered $1$ to $21$. $A$ toy is drawn and then another toy is drawn without replacement. The probability that both toys will show even numbers is

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