Here is the question I didn't get in my last PR test:

**Q. A man chooses an outfit from 3 different shirts, 2 different pairs 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 x (1/2)^3**

**b) (1/3)^6 x (1/2)**

**c) (1/3)^4**

**d) (1/3)^2 x (1/2)**

**e) (1/3)^2 x 5**

**Solution**:

I came up with an extremely simplified, but slightly longer solution, mainly because I didn't "get" the solutions posted completely. The method is still the same, but this is my explanation for why it is the way it is:

Suppose the shirts are - Y (yellow), G (green), W (white)

The shoes are - T (teal), V (violet)

And the pants are - R (red), L (lemon), A (aqua)

Yes, the man has a horrible sense of fashion. Let's call him Punk!

*Note: My explanation is based on a simple rule my math tutor taught me in school -- 'and' always translates into 'x' (multiplication) and 'or' always becomes '+' (addition).*

Now, the probability of him picking up a particular shirt, say Y = 1/3

This is the probability of each shirt, and so P(Y) = P(G) = P(W) = 1/3

Same for pants i.e. P(R) = P(L) = P(A) = 1/3

For shoes = P(T) = P(V) = 1/2

For shoes = P(T) = P(V) = 1/2

__Day 1:__

Mr. Punk can choose an outfit (Shirt & Shoes & Pants) by combining any of the above items.

P (choosing a shirt) = P(Y) or P(G) or P(W)

Similarly,

P (choosing a pair of shoes) = P(T) or P(V)

P (choosing a pair of pants) = P(R) or P(L) or P(A)

One of each item has to be picked up at the same time, so we will multiply the above. The equation is:

P (1Shirt, 1Shoe, 1Pant) = [ P(Y) + P(G) + P(W) ] x [ P(T) + P(V) ] x [ P(R) + P(L) + P(A) ]

The RHS (right hand side of the equation) can be put in words:

He must have picked either the Y 'or' G 'or' W shirt AND a T 'or' V shoe AND a R 'or' L 'or' A pant. I simply converted the 'or' to + & the AND to x.

**P (1Shirt, 1Shoe, 1Pant) = (1/3 + 1/3 + 1/3) x (1/2 + 1/2) x (1/3 + 1/3 + 1/3)**

**= 3/3 x 2/2 x 3/3**

**= 1.**

Most people can derive this by common sense, since the probability of him choosing n combination of outfits has to be 1, because although the combos may be different, the probability of him choosing 1 piece of clothing from each item is a 100% (he will not walk out half-naked!). Anyway, common sense is not so common, so I needed to explain this to myself.

__Day 2__:

Let us suppose that Mr. Punk picked out Y shirt, T shoes & R pants.

Here is where I had made a mistake. I had assumed that the items he could choose his day 2 outfits from had reduced. In reality, he once again had the same choice from 3 shirts, 2 shoes & 3 pants (yes, he doesn't send his clothes to the laundry after wearing them just once!)

Now the limitation of him choosing a different shirt & pants, and the same shoes, is what we have to look for i.e. what is the probability of that happening by chance, not by him deliberately choosing only from the remaining 2 shirts, etc.

The equation for day 2 will look like this:

P (1Shirt, 1Shoe, 1Pant) = [ P(G) + P(W) ] x [ P(T) ] x [ P(L) + P(A) ]

(since he's already worn the Y & R, and he has to choose only T again.)

**P (1Shirt, 1Shoe, 1Pant) = (1/3 + 1/3) x (1/2) x (1/3 + 1/3)**

**= 2/3 x 1/2 x 2/3**

**= 2/9**

__Day 3:__

Mr. Punk wore G shirt, T shoes & L pants on day 2.

For day 3, what is the probability of him wearing W shirt, T shoes & A pants?

**P (W Shirt, T Shoe, A Pant) = P(W) x P(T) x P(A)**

**=**

**1/3 x 1/2 x 1/3**

**= 1/18**

Multiplying the final

**blue**equations for days 1 AND 2 AND 3 (remember, AND => multiplication),

**P (different shirt, pants, and same shoes, on the 3 days) = 1 x 2/9 x 1/18**

**= 1/81 or (1/3)^4**

PS. I've kept my earlier thoughts on the question...just in case I need them some day:

I made one MAIN mistake. I was thinking he had to choose a particular outfit, and I calculated the probability of that. That's why I was getting answer B I think. Anyway, after reading a couple of solutions posted on the web, I figured it out. We're looking for the probability of him not repeating that same outfit, and there IS replacement here! I had assumed that once shirt#1 is gone, he's left to choose between any of the remaining 2, when in fact he still has his 3 shirts to choose from, but now he can actually pick any 1 of the 2. This explanation is beginning to suck.

After some thought, I finally understood the solution. The solutions I found on GMAT forums helped.

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