Challenge Problem 6/27/05 - Hand in Glove

[marc.gagnon]

Problem Statement
06/27/05
Question
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?

Solution Statement
The simplest way to approach a complex probability problem is not always the direct way. In order to solve this problem directly, we would have to calculate the probabilities of all the different ways we could get two opposite-handed, same-colored gloves in three picks. A considerably less taxing approach is to calculate the probability of NOT getting two such gloves and subtracting that number from 1 (remember that the probability of an event occurring plus the probability of it NOT occurring must equal 1).

Let's start with an assumption that the first glove we pick is blue. The hand of the first glove is not important; it could be either right or left. So our first pick is any blue. Since there are 3 pairs of blue gloves and 10 gloves total, the probability of selecting a blue glove first is 6/10.

Let's say our second pick is the same hand in blue. Since there are now 2 blue gloves of the same hand out of the 9 remaining gloves, the probability of selecting such a glove is 2/9.

Our third pick could either be the same hand in blue again or any green. Since there is now 1 blue glove of the same hand and 4 green gloves among the 8 remaining gloves, the probability of such a pick is (1 + 4)/8 or 5/8.

The total probability for this scenario is the product of these three individual probabilities: 6/10 x 2/9 x 5/8 = 60/720.

We can summarize this in a chart:

We can summarize this in a chart:
--------------------------------------------
Pick | Color | Hand Probability |
1st | blue/any | 6/10 |
--------------------------------------------
2nd | blue/same | 2/9 |
--------------------------------------------
3rd | blue/same |
| or any green | 5/8 |
--------------------------------------------
total 6/10 x 2/9 x 5/8 = 60/720


My question
I understand pick1 and 3, however, why are we limiting the second pic to only blue of the same hand, instead of looking at any nonmatching glove (as we do in pick 3).
My thinking would be to do the following:
(Pick1) 6/10 for the selection of any blue glove
(Pick2) 6/9 for any glove that would not create a pair, e.g. same-hand blue, or any non-blue
(Pick3) 5/8 for any glove that would not create a pair, e.g. same-hand blue, or any non-blue

[messi10]

Hi Marc,

If you do not fix pick 2, how will you know what to expect in pick 3?

Lets designate shorthand notations to the gloves we have:
Blue-Left = BL
Blue-Right = BR
Green-Left = GL
Green-Right = GR

Lets consider 2 scenarios:

Scenario 1
Pick 1: BL (Since it was any Blue glove)
Pick 2: GR (Any glove except a BR)

For pick 3, we cannot have a BR or a GL. We are left with 2 BLs and 1 GR. So probability of Pick 3 is 3/8

Scenario 2
Pick 1: BL (Since it was any Blue glove)
Pick 2: BL (Any glove except a BR)

For pick 3, we cannot have a BR. We are left with 1 BL, 2GRs and 2GLs. So probability of Pick 3 is 5/8

So, we need to know the 2nd pick to know the probability for the 3rd pick

Regards

Sunil

[marc.gagnon]

Great Thanks! I wasn't considering the chances of the second pick pairing with the third.

[jnelson0612]

Great!