Probability
Aptitude

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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

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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, 

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

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


(4) Comment(s)


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 Laugh