Problem Solving

A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

Answer: C

Source: www.GMATPrepNow.com
Difficulty Level: 700

Brent@GMATPrepNow wrote: ↑

Fri May 26, 2017 11:11 am

A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

Answer: C

Source: www.GMATPrepNow.com
Difficulty Level: 700

Solution:

Instead of finding the probability that Ed will have at least one pair of matching socks, let’s find the probability that he will have no pairs of matching socks. After all, the former probability is the latter probability subtracted from 1.

Since he has 8 pairs of socks, he has 16 socks. Therefore, he has 16C6 ways to choose 6 socks from 16. Let’s say each pair of socks is of a different color. In other words, there are 8 colors and 2 socks of each color. There are 8C6 ways to choose 6 colors from 8 and for each color there are 2 ways to choose a sock of that color. Therefore, there are 8C6 x 2 x 2 x 2 x 2 x 2 x 2 = 8C6 x 2^6 ways to 6 socks of different colors (i.e., no socks of matching color). Therefore, the probability that he will have no pairs of matching socks is:

8C6 x 2^6 / 14C6 = 28 x 64 / 8008 = 28 x 8 / 1001 = 4 x 8 / 143 = 32/143

Finally, the probability that he will have at least one pair of matching socks is 1 - 32/143 = 111/143.

Alternate Solution:

Instead of finding the probability that Ed will have at least one pair of matching socks, let’s find the probability that he will have no pairs of matching socks and then subtract that probability from 1.

Our approach is to use the multiplication rule for probabilities when we are drawing without replacement.

The first draw has probability 16/16 = 1. For the second draw, there are 15 socks remaining, and he can choose any sock except the match to the first sock, so there are 14 available socks, and the probability is 14/15. For the third draw, there are 14 socks remaining and he can pick any of 12 socks to avoid a match, so the probability is 12/14. Similarly, for the fourth draw, there are 13 remaining socks, and he can pick any of 10 socks to avoid a match, so the probability is 10/13. Notice that for each pick, the number of socks available to be chosen decreases by 1, but the number of socks available to pick, to avoid a match, decreases by 2. Thus, we have:

P(no matches in 6 picks) = 16/16 x 14/15 x 12/14 x 10/13 x 8/12 x 6/11

After canceling and simplifying, we have:

P(no matches in 6 picks) = 32/143

Thus, the probability of at least one match = 1 - 32/143 = 111/143.

Answer: C

Brent@GMATPrepNow wrote: ↑

Fri May 26, 2017 11:11 am

A drawer contains 8 pairs of socks. For each sock, there is exactly one matching sock. If Ed randomly selects 6 socks without replacement, what is the probability that he will have at least one pair of matching socks?

A) 1/4
B) 19/51
C) 111/143
D) 3/4
E) 145/189

Answer: C

Source: www.GMATPrepNow.com
Difficulty Level: 700

Here's an approach that uses probability rules.

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(at least 1 matching pair) = 1 - P(zero matching pairs)

P(zero matching pairs)
P(zero matching pairs) = P(pick ANY sock 1st AND 2nd sock doesn't match 1st sock AND 3rd sock doesn't match other selections AND 4th sock doesn't match other selections AND 5th sock doesn't match other selections AND 6th sock doesn't match other selections)
= P(pick ANY sock 1st) x P(2nd sock doesn't match 1st sock) x P(3rd sock doesn't match other selections) x P(4th sock doesn't match other selections) x P(5th sock doesn't match other selections) x P(6th sock doesn't match other selections)
= 1 x 14/15 x 12/14 x 10/13 x 8/12 x 6/11
= 32/143

So, P(win at least 1 prize) = 1 - P(win zero prizes)
= 1 - 32/143
= 111/143

Answer: C

ASIDE: Once we draw the first sock, there are 15 sock s remaining, and only 1 matches the 1st selection.
This means there are 14 sock s that DON'T match the first.
So, P(no match on 2nd draw) = 14/15

Then, once we draw the second sock (without a match), there are 14 socks remaining, and 2 of them match either the 1st or 2nd selection.
This means there are 12 socks that DON'T match.
So, P(no match on 3rd draw) = 12/14

etc

Cheers,
Brent