Day 12: Picking a Christmas Card

Probability Level 2

Rachel has found that her class' Christmas cards have all been mixed up in a pile. There are 33 cards in the pile, and she knows that there are n n cards that are addressed to her.

She picks two cards at random (without replacement). The probability that both cards are addressed to her is 1 8 \frac18 .

Find the value of n n .

This problem is part of the Advent Calendar 2015 .


The answer is 12.

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Michael Ng by Michael Ng , 22, United Kingdom

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1 solution

Michael Ng
Dec 11, 2015

The probability is n ( n − 1 ) 33 × 32 \frac{n(n-1)}{33\times 32} as the probability is conditional. Therefore n ( n − 1 ) 33 × 32 = 1 8 n ( n − 1 ) − 132 = 0 ( n − 12 ) ( n + 11 ) = 0 \frac{n(n-1)}{33\times 32} = \frac18 \\ n(n-1)-132 = 0 \\ (n-12)(n+11) = 0 So the answer is 12 \boxed{12} as required.

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Can you please elaborate!

Tanvir Hasan - 5 years, 4 months ago

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Yes, definitely. When trying to pick the first card she has n 33 \frac{n}{33} chance of getting a card addressed to her. Now having taken this card, there are n − 1 n-1 cards left addressed to her in the pile, and there are 32 32 cards left over. Now she has a n − 1 32 \frac{n-1}{32} chance of getting a card addressed to her. So therefore she has the probability as mentioned above of getting two cards with her name on them.

Now equate it to 1 8 \frac18 and you can solve the quadratic equation and then you are done.

Michael Ng - 5 years, 4 months ago

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