BTGmoderatorDC wrote:Tony owns six unique matched pairs of socks. All twelve socks are kept loose and unpaired in a drawer. If Tony pulls socks at random, how many must he pull in order to have better than a 50% chance of getting two socks that match?
A) 3
B) 4
C) 5
D) 6
E) 7
We can PLUG IN THE ANSWERS, which represent the minimum number of socks that must be pulled.
When the correct answer choice is plugged in, the probability of NOT picking a matching pair will be LESS than 1/2 (implying that the probability of picking a matching pair will be MORE than 1/2).
Since we need to determine the minimum number of socks that must be pulled, we should start with the SMALLEST answer choice.
Note:
The first sock pulled can be ANY of the 12 socks and thus is irrelevant.
Our only concern is whether any of the SUBSEQUENT socks form a matching pair.
A: 3 socks pulled
After the 1st sock is pulled:
P(2nd sock does not match the 1st) = 10/11. (Of the 11 socks left, 10 do not match the 1st sock.)
P(3rd sock does not match the 1st or 2nd) = 8/10. (Of the 10 socks left, 8 do not match the 1st or 2nd.)
To combine these probabilities, we multiply:
10/11 * 8/10 = 8/11.
Since the resulting probability is not less than 1/2, eliminate A.
B: 4 socks pulled
After the 1st sock is pulled:
P(2nd sock does not match the 1st) = 10/11. (Of the 11 socks left, 10 do not match the 1st sock.)
P(3rd sock does not match the 1st or 2nd) = 8/10. (Of the 10 socks left, 8 do not match the 1st or 2nd.)
P(4th sock does not match 1st, 2nd, or 3rd) = 6/9. (Of the 9 socks left, 6 do not match the 1st, 2nd or 3rd.)
To combine these probabilities, we multiply:
10/11 * 8/10 * 6/9 = 16/33.
Success!
The resulting probability is less than 1/2.
The correct answer is
B.
The OA implies the following:
P(not matching set) = 16/33.
P(matching set) = 1 - 16/33 =
17/33.
The probability in blue is greater than 50%.
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