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 ?
Thanks for the comment. I really appreciate it.
Here is where I think you went wrong. Firstly, you cannot add them all because we are finding the probability of 3 days simultaneously, therefore, we need to multiply them.
Here is the process:
first day
1 way of choosing shoe
3 ways of choosing shirt
3 ways of chossing pant
1x3x3 = 9 different ways
total combinations for the day = 2*3*3 = 18
Probability for the first day = 9/18
second day
1 way of choosing shoe
2 ways of choosing shirt
2 ways of chossing pant
1x2x2 = 4 different ways
total combinations for the day = 2*3*3 = 18
Probability for the second day day = 4/18
third and final day
1 way of choosing shoe
1 way of choosing shirt
1 way of chossing pant
1x1x1 = 1 way
total combinations for the day = 2*3*3 = 18
Probability for the thrid and final day = 1/18
total probability = 9/18 * 4/18 * 1/18 = 1/9 * 1/18
but since we have two shoes:
1/9 * 1/18 * 2 = 1/81
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.
No rest for the Wicked....
