A man chooses an outfit from 3 different shirts, 2 different

Question

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?

1) \((1/3)^6*(1/2)^3\)

2) \((1/3)^6*(1/2)\)

3) \((1/3)^4\)

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

5) \(5*(1/3)^2\)

Happy solving!

A.  (\frac{1}{3})^6*(\frac{1}{2})^3

B.  (\frac{1}{3})^6*(\frac{1}{2})

C.  (\frac{1}{3})^4

D.  (\frac{1}{3})^2*(\frac{1}{2})

E.  5*(\frac{1}{3})^2

Happy solving!

Bunuel · Accepted Answer

For the first day he can choose  any  outfit,  p=1 ;

For the second day he must choose the same shoes as on the first day and different shirts and pants form the first day's,  p=\frac{1}{2}*\frac{2}{3}*\frac{2}{3}=\frac{2}{9} ;

For the third day he must choose the same shoes as on the first day and different shirts and pants from the first and second day's,  p=\frac{1}{2}*\frac{1}{3}*\frac{1}{3}=\frac{1}{18} ;

P=1*\frac{2}{9}*\frac{1}{18}=\frac{1}{81}=\frac{1}{3^4}

Answer: C ( \frac{1}{3^4} ).

sriharimurthy · Answer

Having Bunuel post is as good as posting the OA... :wink:

In fact, better.. because not only are the answers always right but the explanations always perfect! +1 to you Bunuel! (I think soon you'll have more kudos than posts!) Cheers!

h2polo · Answer

Good problem.

The first day the man can wear anything:
(Shoes)*(Shirts)*(Pants)
(2/2)*(3/3)*(3/3)

The second day the man must wear the same pair of shoes and not wear the shirt and pants from the day before:
(1/2)*(2/3)*(2/3)

The final day, the man must continue to wear the same pair of shoes and wear the one shirt and one pants that were not worn:
(1/2)*(1/3)*(1/3)

So multiplay all together to get the probability of this happening each consecutive day:

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

ANSWER: C.

ctrlaltdel · Answer

h2polo,

Kudos to you as well, u have been ruling the quant section, soon i see u becoming the moderator.
But keep in mind, u r competing with the best....Bunuel.

h2polo · Answer

Oh no... there is no competing with Bunuel. He solves the unsolvable. Hahaha...

pleonasm · Answer

I missed out on the activities in this forum lately and it's again a joy to read Bunuel's answers. Kudos

san03 · Answer

My question is:

day 2: probability of choosing shoe should be 1 as we can not choose one of the two shoes rather the one that was already chosen on day 1
same for day 3
please explain..

nickk · Answer

san03: On day 2 (and 3), the probability for the shoes should be 1/2
This is because given a shoe was chosen on day 1, the probability of choosing that same shoe the next day is 1/2.

san03 · Answer

Aah!! i think I got confused between number of ways and probability.
probability of choosing same shoe= 1C1/2C1
Many thanks

ShashankDave · Answer

Another way to look at the problem :-
!. Since a single pair of shoes has to be chosen and worn for all the three days,
No. of ways of selecting one pair of shoes from the available two = 2C1 = 2.
2. Now the remaining task for us is to choose different shirts and pants for the three days : -
1st day = 3 X 3
2nd day = 2 X 2
3rd day = 1 X 1
Multiplying all the terms
2C1 X 2 X 2 X 3 X 3 X 1 X 1 (Numerator)

now what will come in the denominator?
day Pants Pair of shoes Shirts
1st 3 2 3
2nd 3 2 3
3rd 3 2 3

That is = 3^3 X 2^3 X 3^3 (Denominator)

When the Numerator is divided by the Denominator, you will get the required answer = 1/(3)^4.
Kudos for the answer! :-D

justbequiet · Answer

1st Day -  \frac{3}{3}  shirts  *   \frac{2}{2}  shoes  *   \frac{3}{3}  pants

2nd Day -  \frac{2}{3}  shirts,  *   \frac{1}{2}  shoes  *   \frac{2}{3}  pants

3rd Day -  \frac{1}{3}  shirts,  *   \frac{1}{2}  shoes  *   \frac{1}{3}  pants

Answer is  (\frac{1}{3})^4

sankalp1992 · Answer

the probability of choosing a unique set of attire on the fisrt day = 1.
from the second day, if I choose to wear the same pair of shoes(s1) on second and third day, then the probaility =  * = 1/81.

Similarly,I can choose to wear shoes(S2) on all three days. Here as well the probability = 1/81.

Hence, the total probaility = 1/81 + 1/81=  2/81

shashankism · Answer

3 different shirts, 2 different pair of shoes, and 3 different pants..
For first morning, Probability of selecting 1 shirt, 1 pair of shoes, and 1 pair of pants = 3/3 * 2/2 *3/3 =1 
For 2nd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 2/3 * 1/2 * 2/3 = 2/9 (2 fresh shirts left and 2 fresh pair of pants left)
For 3rd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 1/3 * 1/2 * 1/3 = 1/18 (1 fresh shirt left and 1 fresh pair of pants left)

So, probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated = 1 * 2/9 * 1/18 =  \frac {1}{3^4}

Answer C

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