Aptitude 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,

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

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

**Mani**

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**JPyeah**

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

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

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**

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it's amazing it helps me so much

3/14 is the correct answer