Aptitude Discussion

Q. |
Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King. |

✖ A. |
1/13 |

✖ B. |
4/13 |

✖ C. |
1/52 |

✔ D. |
1/26 |

**Solution:**

Option(**D**) is correct

Let $X$ be the event that cards are in a club which is not king and other is the king of club.

Let $Y$ be the event that one is any club card and other is a non-club king.

Hence, required probability:

$= P(A) + P(B)$

$= \dfrac{ {^{12}C_1} × {^1C_1} }{^{52}C_2} + \dfrac{ {^{13}C_1} × {^3C_1} }{^{52}C_2}$

$=\left( \dfrac{2×(12×1)}{52×51} \right) + \left(\dfrac{2(13×3)}{52×51}\right)$

$= \left(\dfrac{24+78}{52×51}\right)$

$= \textbf{1/26}$

**Edit:** For explanation on the given solution, read comment by **Saurabh Bansal**

**Hoor**

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Say the first card is a club and the second card is a king. however if the first card is king of clubs then you have only three kings left for the second card.

So for clarity, split the set of club cards into two sets: the first set of 12 cards except the king of clubs and the second set has the just king of clubs.

Now read the solution again, it will be clearer now ☺

**Sandip**

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I think the answer is 1/26.

As Question doesn't enforces that club can't have the king.

So for 12 clubs, any 4 kings can be selected. and When a king club is selected you have 3 kings to be select from

$12*4+1*3=51$

So, the answer is,

$\dfrac{51}{{^{52}C_2}} = \dfrac{51}{(52*51/2)} = \dfrac{1}{26}$

$hy ${^{52}C_2}$ at the denominator?

It should be $52*51$

**Neena**

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How did you arrive the below calculation as;

$\dfrac{2 \times (12 \times 1)}{52 \times 51}+\dfrac{2(13 \times 3)}{52 \times 51}$

I don't understand from where '2'came in this calculation. Kindly clarify.

I got the final answer as $\dfrac{1}{52}$.

BECAUSE ANYONE CAN COME FIRST.

2 is came from solve $\dfrac{52*51}{2*1}$

**Tanmoy**

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i can't understand this......

**Avantika**

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the total number of kings are 4.

So, the number of ways a king can be selected is ${^4C_1}$ and the total number of clubs are 12 excluding the king club.

So, the probability should be $\dfrac{{^4C_1} \times {^{12}C_1}}{{^{52}C_2}}$

Am I correct?

**Ajith Sanjay**

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i cant understand the solution.

**Karthi**

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I believe that the final answer is $\dfrac{1}{52}$

12.4/52c2+1.3/52c2=1/26

it is 52c2 coz order of the cards doesn't matter

I didn't understand this solution somebody can explane it in some easy way?