Aptitude Discussion

Q. |
Four cards are drawn at random from a pack of 52 plating cards. Find the probability of getting all the four cards of the same suit. |

✖ A. |
13/270725 |

✖ B. |
91/190 |

✖ C. |
178/20825 |

✔ D. |
44/4165 |

**Solution:**

Option(**D**) is correct

Four cards can be selected from 52 cards in ${^{52}C_4}$ ways.

Now, there are four suits, e.g. club, spade, heart and diamond each of 13 cards.

So total number of ways of getting all the four cards of the same suit:

$⇒ {^{13}C_4} + {^{13}C_4} + {^{13}C_4} + {^{13}C_4}$

$= 4 × {^{13}C_4}$

So required probability,

$= \dfrac{4 × {^{13}C_4}}{^{52}C_4}$

$= \textbf{44/4165}$

**Edit:** Final answer has been changed from 198/20825 to 44/4165, after recieving a comment from **Saurabh Bansal.**

**Saurabh Bansal**

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Thank you Saurabh for letting me know, corrected the mistake.

**Btm**

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The solution is correct but the answer they provide is wrong (maybe a calculation error).

using software (R):

4 * choose(13, 4)/ choose(52, 4) = 44 /4165 or 132/12495

**Steve**

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I keep comming up with $4C1*13C4/52C4$ to be $220/20825$ which is exactly the same as subodh and danh

**Subodh**

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Answer is

$(4\times 13C4)/(52C4) = 132/12495 = 44/4165$

**Danh**

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The correct probability is 4 times the probability of four cards of one suit, which is four times $13/52 * 12/51 * 11/50 * 10/49$.

So, the correct answer is,

$= 4 * 13/52 * 12/51 * 11/50 * 10/49$

$= 12 * 11 / 51 / 49 / 5$

$= (12 * 11) / (51 * 49 * 5)$

$= 132/12495$.

Mohammed Javad is correct.

The final computation is wrong.

It should be,

$\dfrac{4 \times ^{13}C_4}{^{52}C_4}=\dfrac{44}{4165}$