If the probability of pulling two cards from a stack of uniquely numbe

Hi All,

This question can be solved by TESTing THE ANSWERS.

We're told that the probability of pulling two cards from a stack of uniquely numbered cards (without replacement) and getting the '5' and '6' is 0.10... We're asked for the total number of cards in the stack.

To start, it's worth noting that .1 is a relatively 'nice' decimal. Probability questions almost always involve multiplying fractions - and the results are often NOT 'nice' decimals. This means that we're likely dealing with a total that can be converted into 'nice' fractions (for example 1/4 is 'nice', while 1/11 is NOT).

Let's TEST Answer B: 4 cards

Pulling the '5' or the '6' on the first try = 2/4
Pulling the other card on the second try = 1/3
Probability of pulling the two cards we want = (2/4)(1/3) = 2/12 = 1/6. This is NOT a match for what we were told (it's supposed to be .1 = 1/10).

Let's TEST Answer C: 5 cards

Pulling the '5' or the '6' on the first try = 2/5
Pulling the other card on the second try = 1/4
Probability of pulling the two cards we want = (2/5)(1/4) = 2/20 = 1/10. This IS an exact match for what we were told, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

We are given that the probability of getting the 5 and the 6 is the probability of getting the 5 followed by the 6 AND the probability of getting the 6 followed by the 5.

If we let n = the number of cards in the deck, the probability of getting the 5 followed by the 6 is 1/n x 1/(n - 1), and this also will be the probability of the 6 followed by the 5. Since we are given that the probability of selecting the 5 and the 6 is 0.1 = 1/10, we can create the following equation:

1/n x 1/(n-1) + 1/n x 1/(n-1) = 1/10

2/(n^2 - n) = 1/10

n^2 - n = 20

n^2 - n - 20 = 0

(n - 5)(n + 4) = 0

n = 5 or n = -4

Since n can’t be negative, n = 5.

Answer: C

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