A bag contains 21 toys numbered 1 to 21 . A toy is drawn and then another toy is drawn without replacement.…
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
$\frac{5}{21}$
$\frac{3}{14}$
$\frac{11}{42}$
$\frac{4}{21}$
Solution
Total number of toys numbered from 1 to 21 .
$
\mathrm{P} \text { (two toys will show even numbers) }==\frac{{ }^{10} \mathrm{c}_2}{{ }^{21} \mathrm{c}_2}
$
The total number of even numbers between 1 to 21 is 10.
$
\text { So, }=\frac{{ }^{10} \mathrm{~L}_2}{{ }^{21} \mathrm{~L}_2}=\frac{10 \times 9}{21 \times 20}=\frac{8}{14} \text {. }
$