# Easy Probability Solved QuestionAptitude Discussion

 Q. A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement. Find the probability that both toys will show even numbers.
 ✖ A. 5/21 ✔ B. 9/42 ✖ C. 11/42 ✖ D. 4/21

Solution:
Option(B) is correct

The probability that first toy shows the even number,

$= \dfrac{10}{21}$

Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.

Hence, probability that second toy shows the even number,

$= \dfrac{9}{20}$

Required probability,

$= \left(\dfrac{10}{21}\right)×\left(\dfrac{9}{20}\right)$

$= \dfrac{\textbf{9}}{\textbf{42}}$

## (5) Comment(s)

Mani
()

JPyeah
()

Does it really have to be 9/14 or you can also have 3/14 as the answer

JPyeah
()

Does it really have to be 9/42 or you can also have 3/14 as the answer

Kalpana
()

You sure can have $\dfrac{3}{14}$ as the final answer, but since that's not given in the option choices, I guess $\dfrac{9}{42}$ is ok too.

Funky J
()

it's amazing it helps me so much