BTGmoderatorDC wrote: ↑

Sun Feb 09, 2020 8:13 pm

In packing for a trip, Sarah puts three pairs of socks - one red, one blue, and one green - into one compartment of her suitcase. If she then pulls four individual socks out of the suitcase, simultaneously and at random, what is the probability that she pulls out exactly two matching pairs?

A. 1/5
B. 1/4
C. 1/3
D. 2/3
E. 4/5

OA A

Source: Veritas Prep

No. of ways of selecting any 4 socks out of 6 = 6C4 = 6C2 = (6.5)/(1.2) = 15;
No. of ways of selecting any 2 pair of socks out of 3 = 3C2 = 3C1 = 3

Probability that she pulls out exactly two matching pairs = 3/15 = 1/5

The correct answer: A

Hope this helps!

-Jay
_________________
Manhattan Review

Locations: Manhattan Review Jayanagar | GMAT Prep Tarnaka | GRE Prep Madhapur | Kukatpally GRE Coaching | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

BTGmoderatorDC wrote: ↑

Sun Feb 09, 2020 8:13 pm

In packing for a trip, Sarah puts three pairs of socks - one red, one blue, and one green - into one compartment of her suitcase. If she then pulls four individual socks out of the suitcase, simultaneously and at random, what is the probability that she pulls out exactly two matching pairs?

A. 1/5
B. 1/4
C. 1/3
D. 2/3
E. 4/5



OA A

Source: Veritas Prep

Solution:

We have three scenarios of two matching pairs: 1) a red pair and a blue pair; 2) a red pair and a green pair; 3) a blue pair and a green pair. Let’s start with the probability of selecting a red pair and a blue pair. To select a red pair and a blue pair is to select two red socks and two blue socks. So let’s assume the first two socks are red and the last two socks are blue; the probability of selecting these socks in that order is:

P(R, R, B, B) = 2/6 x 1/5 x 2/4 x 1/3 = 1/6 x 1/5 x 1/3 = 1/90.

However, the two red socks and the two blue socks, in any order, can be selected in 4!/(2! x 2!) = 24/4 = 6 ways. Thus, the probability of two red socks and two blue socks is:

P(2R and 2B) = 1/90 x 6 = 6/90 = 1/15.

Using similar logic, we see that the probability of pulling a red pair and a green pair is 1/15, and so is the probability of pulling a blue pair and a green pair. Thus, the total probability is:

1/15 + 1/15 + 1/15 = 3/15 = 1/5.

Alternate Solution 1:

From a total of 6 socks, two pairs, i.e., 4 socks, can be pulled in 6C4 = 6!/(4! 2!) = (6 x 5)/2 = 3 x 5 = 15 ways.

Three of these choices contain two matching pairs, namely: 1) a red pair and a blue pair, 2) a blue pair and a green pair; 3) a red pair and a green pair.

Therefore, the probability of pulling two matching pairs is 3/15 = 1/5.

Alternate Solution 2:

We note that when two pairs are pulled, one pair is left in the suitcase. Thus, choosing two pairs is equivalent to choosing the pair that is left in the suitcase. We will calculate the probability that the pair that’s left in the suitcase is a matching pair (so that the socks that were pulled must contain two matching pairs).

No matter which sock has been chosen first, there is one matching sock among the remaining five socks; thus the probability that the chosen pair of socks is a matching pair is 1/5.

Answer: A