Probability Tricky Qn

See given info is

3 shirts, 2 shoes, 3 pants

total nos of ways of choosing in a day is 3c1*2c1*3c1..............................(x)

Now lets say he chose a pair of shoes

so for the first day choices he has are
3c1(shirt)*3c1(pant)*1(the shoe that he chose).....(1)

to find probability divide (1) by total (x)

For day 2
2c1(shirt) *2c1(pant) *1

Again for probability...divide by x

for day 3
1c1(shirt) *1c1 (pant) *1

Again for probability...divide by x

Now multiply these 3 variable for 3 days

Multiply the answer by 2 (because if he had chosen the other show at start, there would have been similar cases as well)

when you solve the expression is like this

day1*day2*day3*2

(18/18)*(8/18)*(2/18)*2

=1/81=1/3^4

C it is

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 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 of (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

So the correct answer is C.

Please note that the explanation above might seem long, but the process is quick and easy if you know what to do!

Great Explanation. Clear as a crystal. Thanks for your time and patience.

kvcpk wrote:Great Explanation. Clear as a crystal. Thanks for your time and patience.

Happy to help!

GMATGURU

Thank you so much. These kind of explanations are really really helpful and lots of people like me are craving for this kind of posts. Thank u so much

winnerhere wrote:GMATGURU

Thank you so much. These kind of explanations are really really helpful and lots of people like me are craving for this kind of posts. Thank u so much

Glad you found the explanation helpful! Now the goal is be able to replicate the process of solving should you see a similar problem in the future.