Probability

Anindya Madhudor wrote:A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

a. 3/10
b. 23/60
c. 7/12
d. 41/60
e. 5/6

OA: D

P(matching set) = 1 - P(no matching set).

There are 3 ways to NOT select a matching set of gloves.

Case 1: 3 non-matching blue
P(1st glove is blue) = 6/10.
P(2nd glove is blue and for the same hand) = 2/9.
P(3rd glove is blue and for the same hand) = 1/8.
Since we want all of these events to happen, we multiply the fractions:
6/10 * 2/9 * 1/8 = 1/60.

Case 2: 1 green, 2 non-matching blue
P(1st glove is green) = 4/10.
P(2nd glove is blue) = 6/9.
P(3rd glove is blue and for the same hand as the 2nd glove) = 2/8.
Since we want all of these events to happen, we multiply the fractions:
4/10 * 6/9 * 2/8 = 4/60.
Since the one green glove could be 1st, 2nd or 3rd, we multiply by 3:
4/60 * 3 = 12/60.

Case 3: 1 blue, 2 non-matching green
P(1st glove is blue) = 6/10.
P(2nd glove is green) = 4/9.
P(3rd glove is green and for the same hand as the 2nd glove) = 1/8.
Since we want all of these events to happen, we multiply the fractions:
6/10 * 4/9 * 1/8 = 1/30.
Since the one blue glove could be 1st, 2nd or 3rd, we multiply by 3:
1/30 * 3 = 3/30 = 6/60.

Since any of the 3 cases above would yield no matching set of gloves, we add the probabilities:
P(no matching set) = 1/60 + 12/60 + 6/60 = 19/60.

Thus, P(matching set) = 1 - 19/60 = 41/60.

The correct answer is D.

Thank you, Mitch. You are awesome.

Why this question can't be answered by below method..
Total number of possible combinations = 10C3 = 120.
No of ways to pick 3 such that a pair is included is 5C1*8C1(5C1= selection of a pair from 5, and 8C1= selection of 1 glove from remaining 8)

=80

Thus the probability of selecting a pair would be 80/120 or 40/60..

but still I am missing the exact answer 41/60...
can you please explain where i am actually missing or this isn't the right approach to be followed for solving this question..
Thanks....

viveksingh222 wrote:Why this question can't be answered by below method..
Total number of possible combinations = 10C3 = 120.
No of ways to pick 3 such that a pair is included is 5C1*8C1(5C1= selection of a pair from 5, and 8C1= selection of 1 glove from remaining 8)

=80

Thus the probability of selecting a pair would be 80/120 or 40/60..

but still I am missing the exact answer 41/60...
can you please explain where i am actually missing or this isn't the right approach to be followed for solving this question..
Thanks....

The portion in red assumes that there are 5 DISTINCT pairs of gloves.
Let LB = left-handed blue glove, RB = right-handed blue glove, LG = left-handed green glove, and RG = right-handed green glove.
The 3 pairs of blue gloves and 2 pairs of green gloves are as follows:
LB, LB, LB, RB, RB, RB, LG, LG, RG, RG.
Any LB can be combined with any RB to form a matched set of blue gloves.
Either LG can be combined with either RG to form a matched set of green gloves.
Thus, there are more than 5 ways to choose a matched set of gloves.

hmm..it makes sense now...thanks...

Mitch, i'm trying to use combination technique bt its nt working here. can you explain thru dis method