Binominal question

Explanation of the givenanswer :

To hold the ace of hearts :
• Get the ace of hearts (1C1).
• Get 12 more cards from the rest 51 cards (51C12).
No. of ways to hold the ace of hearts = (1C1) (51C12)

To hold all the aces and the ace of hearts :
• Get the ace of hearts (1C1).
• Get all of the rest 3 aces (3C3).
• Get 9 more cards from the rest 48 cards = (48C9)
No. of ways to hold all the aces and the ace of hearts = (1C1)(3C3) (48C9)

Hence, the required probability
= P([hold all the aces] | [hold the ace of hearts])
= P([hold all the aces] and [hold the ace of hearts] / P(hold the ace ofhearts)
= [(1C1) (3C3) (48C9)]/ [(1C1) (51C12)]
= [(3C3) (48C9)]/ (51C12)


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Amendment of your answer :

The main idea of your answer is alright, but mistakes occurs in calculation P(holdthe ace of hearts).
• It should be 1C1 instead of 4C1. 1C1 means that the onlyone cards of the ace of hearts is got, but 4C1 means thatone of the four aces is got.
• It should be 51C12 instead of 48C12. 51C12 means that 12 cards are got from the 51 cards thatare not the ace of hearts, but 48C12­ means that 12 cardsare got from the 48 cards that are not aces.

Hence, your answer would be amended as :
The required probability
= P([hold all the aces] | [hold the ace of hearts])
= P([hold all the aces] / P(hold the ace of hearts)
= [(4C4) (48C9)]/ [(1C1) (51C12)]


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The two answers are equal :

The two answers above are equal to each other, i.e.
[(3C3) (48C9)] / (51C12) = [(4C4) (48C9)] / [(1C1) (51C12)]

This is because : 1C1 = 3C3 = 4C4= 1