The question is: 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 * (1/2)^3 B (1/3)^6 * (1/2) C (1/3)^4 D (1/3)^2 * (1/2) E 5*(1/3)^2 :

On the first day, man has to wear 1 shirt from 3 shirts, 1 pant from 3 pants and 1 shoe from 2 shoes 3C1/3*3C1/3*2C1/2 On the second day, man has to wear 1 shirt of the remaining 2 shirts, 1 pant from the remaining 2 pants and the same shoe 2C1/3*2C1/3*1C1/2 On the third day, man has to wear 1 shirt of the remaining 1 shirt, 1 pant from the remaining 1 pant and the same shoe 1C1/3*1C1/3*1C1/2 So probability for all 3 days is: (1*1*1)*(2/3*2/3*1/2)*(1/3*1/3*1/2) = (1/3)^4 Hope that helps! https://goal-mba.blocked

That's great, the OA = C. But Y you keep dividing by 3 and 2 on the second and third day?

No of ways of selecting a pant/shirt as per the case __________________________________________ The total no of possibilities The denomitor is always the no of options available that is 2 if case is shoes else 3. The numerator varies depnding on the case, the particular selection. https://goal-mba.blocked

possible outcomes in favour of the case/ total outcomes

Well Madhavi, On the second day, as the shirt and the pant is not re-used, the possible outcomes reduces to 2 for the pants and shirts but remains 2 for the shoes and on the third day reduces to 1. Isn't it : Please Help, I am a bit rusted on Prob. and Combinatorics.

Lets think of a six faced dice. Possibility of getting a 2. How do you handle this? You work out the options for this in the numerator and the denominator has all the possible outcomes. Right? Similarly, on the second day you have 3 shirts to select from. But you want to check for the event where you would not want to repeat a shirt. Hence, 3 in the denominator