Probability Question !

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Mani_mba: Master | Next Rank: 500 Posts; Posts: 267; Joined: Wed Jul 16, 2008 12:20 amFollowed by: 1 member; Thanked: 4 times

by Mani_mba » Mon Oct 27, 2008 3:21 am

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.

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

Re: Probability Question !

by parallel_chase » Mon Oct 27, 2008 4:11 am

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?

No rest for the Wicked....

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KeyserSoze525: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Wed Aug 27, 2008 6:37 pm; Thanked: 3 times

by KeyserSoze525 » Mon Oct 27, 2008 4:24 am

What are the possible answer choices?

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mental: Master | Next Rank: 500 Posts; Posts: 145; Joined: Mon Sep 29, 2008 1:14 am; Thanked: 13 times

by mental » Thu Nov 06, 2008 7:31 am

OA?..................is it 1/81

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Mani_mba: Master | Next Rank: 500 Posts; Posts: 267; Joined: Wed Jul 16, 2008 12:20 amFollowed by: 1 member; Thanked: 4 times

by Mani_mba » Thu Nov 06, 2008 9:54 am

Yes, the OA is 1/81.

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

Re: Probability Question !

by logitech » Thu Nov 06, 2008 5:25 pm

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 ?

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members; Thanked: 146 times

Re: Probability Question !

by parallel_chase » Fri Nov 07, 2008 4:03 am

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....

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

Re: Probability Question !

by logitech » Fri Nov 07, 2008 9:53 am

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!

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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x2suresh: Master | Next Rank: 500 Posts; Posts: 258; Joined: Thu Aug 07, 2008 5:32 am; Thanked: 16 times

Re: Probability Question !

by x2suresh » Thu Feb 19, 2009 8:49 pm

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.

1* (2/3*1/2*2/3)*(1/3*1/2*1/3) = 1*2/9* 1/18 = 1/81

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