Mani_mba wrote: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?

OA after discussions !

Thanks.

parallel_chase wrote:

Mani_mba wrote: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?

OA after discussions !

Thanks.

probability of choosing any shoe = 1/2
probability of wearing same shoe for straight 3 days = (1/2)^3
but we have 2 pair of shoes, therefore, (1/2)^3 + (1/2)^3 = 1/8+1/8 = 1/4

Probability of wearing different pant for 3 days = 3/3*2/3*1/3 = 2/9

Probability of wearing different shirt for 3 days = 2/9

total probability = 2/9 * 2/9 * 1/4 = 1/81


OA?

logitech wrote:

parallel_chase wrote:

Mani_mba wrote: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?

OA after discussions !

Thanks.

probability of choosing any shoe = 1/2
probability of wearing same shoe for straight 3 days = (1/2)^3
but we have 2 pair of shoes, therefore, (1/2)^3 + (1/2)^3 = 1/8+1/8 = 1/4

Probability of wearing different pant for 3 days = 3/3*2/3*1/3 = 2/9

Probability of wearing different shirt for 3 days = 2/9

total probability = 2/9 * 2/9 * 1/4 = 1/81


OA?

Parallel Case - you are the man!

But here is where I get confused:

How come you do not think each day separately ?

I was trying to think every day probabilities and add them up.

OR for example

first day

1 way of choosing shoe
3 ways of choosing shirt
3 ways of chossing pant

1x3x3 = 9 different ways

second day

1 way of choosing shoe
2 ways of choosing shirt
2 ways of chossing pant

1x2x2 = 4 different ways

third and final day

1 way of choosing shoe
1 way of choosing shirt
1 way of chossing pant

1x1x1 = 1 way

so 9+4+1 = 14 way

but since we have two shoes:

14x2 = 28 different ways

now if we think three days

2 way of choosing shoe
3 ways of choosing shirt
3 ways of chossing pant

2x3x3 = 18 ways

for 3 days = 18x4=54 ways

28/54 = 14/27 almost 50 % - WHICH SOUNDS TERRIBLY WRONG but where do I make mistake ?

parallel_chase wrote:

logitech wrote:

parallel_chase wrote:

Mani_mba wrote:
I am sure you must have realized it by now where exactly you went wrong.


Instead of 18*3 it should be 18^3 for the total combinations.

Instead of (9+4+1) it should be 9*4*1*2 for the favorable combinations.


I hope it is clear. If you still have any doubts pls do let me know.

You are an artists! THANK YOU!

Mani_mba wrote: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?

OA after discussions !

Thanks.