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by yellowho » Mon Feb 21, 2011 3:23 am
A man chooses an outfit from 3 different shirts, 2 different pair of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?
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by GMATGuruNY » Mon Feb 21, 2011 3:42 am
yellowho wrote:A man chooses an outfit from 3 different shirts, 2 different pair of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

A. (1/3)^6 * (1/2)^3
B. (1/3)^6 * (1/2)
C. (1/3)^4
D. (1/3)^2 * (1/2)
E. 5 * (1/3)^2
The first day is immaterial; the man can wear whatever he wants. We're concerned only with what the man chooses to wear the second and third days. Here are the outcomes that we need:

Second day: different shirt, different pair of pants, same pair of shoes
Third day: different shirt from those worn on days 1 and 2, different pair of pants from those worn on days 1 and 2, same pair of shoes


Let's determine each probability separately:

Probability of choosing a different shirt on the second day: 2/3 (because he has 3 shirts total, and he can't wear the same shirt worn the first day, leaving him 2 good choices)

Probability of choosing a different pair of pants on the second day: 2/3 (because he has 3 pairs total, and he can't wear the same pair worn the first day, leaving him 2 good choices)

Probability of choosing the same pair of shoes: 1/2 (because he has 2 pairs total, and he has to wear the same pair worn the first day, leaving him 1 good choice)

Probability of choosing a different shirt on the third day: 1/3 (because he has 3 shirts total, and he can't wear the same shirts worn the first and second days, leaving him only 1 good choice)

Probability of choosing a different pair of pants on the third day: 1/3 (because he has 3 pairs total, and he can't wear the same pairs worn the first and second days, leaving him only 1 good choice)

Probability of choosing the same pair of shoes: 1/2 (because he has 2 pairs total, and he has to wear the same pair worn the first day and second days, leaving him 1 good choice)

We need all of these events to happen together in order to get a good outcome. To determine the probability that multiple events will happen together, remember this rule:

Probability (A and B) = Probability (A) * Probability (B)

We multiply the probabilities because the more things we want to happen together, the smaller the probability, and when you multiply fractions, the result keeps getting smaller.

So let's take the probabilities that we determined above and multiply:

2/3 * 2/3 * 1/2 * 1/3 * 1/3 * 1/2 = (1/3)^4

The correct answer is C.
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by Anurag@Gurome » Tue Feb 22, 2011 8:54 pm
yellowho wrote:A man chooses an outfit from 3 different shirts, 2 different pair of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

Solution:
Required probability is = (Total number of favorable outcomes)/(Total number of all possible outcomes).
Let us calculate the total number of all possible outcomes.
From 3 different shirts, 2 different pair of shoes and 3 different pants, a triplet of (shirt, shoes, pants) can be formed in 3*2*3 = 18 ways.
So, any one of these 18 combinations can be used on any of the 3 days. Also there can be repetitions.
Hence, total number of all possible outcomes is 18*18*18.
Next, we calculate the number of favorable outcomes.
Suppose, the man wears the first pair of shoes everyday.
Now, 3 different shirts can be worn on the three days in 3! ways because there has to be no repetition.
Similarly 3 different pants can be worn on the three days in 3! ways, again because there has to be no repetition.
So, with first pair of shoes on all 3 days, there can be3! * 3! or 36 ways of wearing different combinations such that the pants and shirts are not repeated.
Likewise, with second pair of shoes on all three days, we can have 36 combinations such that the pants and shirts are not repeated.
So, total number of favorable cases is 36+36 = 72.
Or required probability is 72/(18*18*18) = 1/81.
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