Aptitude Discussion

Q. |
From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack? |

✖ A. |
19/5525 |

✖ B. |
21/5525 |

✖ C. |
17/5525 |

✔ D. |
16/5525 |

**Solution:**

Option(**D**) is correct

Required probability:

\(=\dfrac{^4C_1\times ^4C_1\times^4C_1 }{^{52}C_3}\)

\(=\dfrac{4\times 4\times 4}{22100}\)

\(=\dfrac{16}{5525}\)

**Ayush**

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**Crystal**

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here we need to take all the possible pairs in which event can occur ..i.e we need to take all the ordered pair..

$\dfrac{6! \times 4 \times 4\times 4}{^{52}C_3}$

Hey Cristal,

How have you arrived at multiplying factor $6!$?

Is it related to calculation of ordered pair that you have mentioned in your Comment?

**Anil Kumar**

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Mukesh Ji,

You are considering only one case. There are 5 more cases.

**Anil Kumar**

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J you are correct which is same as the solution.

$48/16575 = 16/5525$

You can cancel num and denom by 3

**J**

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This doesn't seem right to me ....

I have :

$12/52 * 8/51 * 4/50 = 48/16575$

12 choices for the first draw

8 since the four are no longer eligible

4 since the eight are no longer eligible

Since each draw reduces the possibilities by one, denom is $52*51*50$

When I saw this numerical, I also felt same as "crystal" feels

But here there is NO NEED to deal with arrangements (3!=6) because we just want an ace ,a jack and a queenew

In which order we get them is not important

So

Multiplying by 6 is NOT NECESSARY