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no other piece of clothing is repeated?

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no other piece of clothing is repeated?

by bhumika.k.shah » Sun Jan 31, 2010 1:47 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?

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

i tried the 3C1 approach but mid way lost track of it...
how to go about with this sum...different approaches
????
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by bhumika.k.shah » Sun Jan 31, 2010 1:55 am
Why will the probability on the first day be 1 ?
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by ajith » Sun Jan 31, 2010 1:56 am
bhumika.k.shah 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?

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

i tried the 3C1 approach but mid way lost track of it...
how to go about with this sum...different approaches
????
The probability that he wears same pair shoes each day =(1/2)^2 [first day he can wear either and he wears the same next day and the day after = 1*1/2*1/2]
Probability that he wears different pairs of pants every day = 2/9
Probability that he wears different shirts every day = 2/9

Now these events are independent (ie choosing a shirt doesnt depend on what shoes is he wearing or what pants is he wearing)

Hence the probability that all these events occur is a simple multiplication of these = 4/81*1/4 = 1/81 = 1/3^4

IMO, C :)
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by ajith » Sun Jan 31, 2010 1:57 am
bhumika.k.shah wrote:Why will the probability on the first day be 1 ?
Because on first day he has the luxury to wear any of the pairs (he either has to repeat(shoes) or not repeat (pants, shirts) in the subsequent days.
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by bhumika.k.shah » Sun Jan 31, 2010 1:58 am
ajith i dint get it !

please explain step wise with detailed explanations :(
ajith wrote:
bhumika.k.shah 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?

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

i tried the 3C1 approach but mid way lost track of it...
how to go about with this sum...different approaches
????
The probability that he wears same pair shoes each day =(1/2)^2 [first day he can wear either and he wears the same next day and the day after = 1*1/2*1/2]
Probability that he wears different pairs of pants every day = 2/9
Probability that he wears different shirts every day = 2/9

Now these events are independent (ie choosing a shirt doesnt depend on what shoes is he wearing or what pants is he wearing)

Hence the probability that all these events occur is a simple multiplication of these = 4/81*1/4 = 1/81 = 1/3^4

IMO, C :)
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by ajith » Sun Jan 31, 2010 2:04 am
The probability that he wears same pair shoes each day =(1/2)^2

[Say he has to pairs A and B now he can wear AAA three days or BBB three days and in total he has 8 different arrangements (2*2*2) so the probability that he wears the same shoes is 2/8 (2 favourable (AAA, BBB)/ total outcomes)]


Probability that he wears different pairs of pants every day = 2/9

[Say He has 3 Pairs of pants X, Y, Z now the total no of ways of wearing the clothes = 3*3*3

Now out which XYZ, XZY, YXZ, YZX, ZXY, ZYX are favorable so the probability = 6/27 = 2/9]


Probability that he wears different shirts every day = 2/9

[ Explanation same as that of pants]


Hope it helps!
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by bhumika.k.shah » Sun Jan 31, 2010 2:22 am
How can this be done in the formula way ...
like 3C1 and blah!
ajith wrote:The probability that he wears same pair shoes each day =(1/2)^2

[Say he has to pairs A and B now he can wear AAA three days or BBB three days and in total he has 8 different arrangements (2*2*2) so the probability that he wears the same shoes is 2/8 (2 favourable (AAA, BBB)/ total outcomes)]


Probability that he wears different pairs of pants every day = 2/9

[Say He has 3 Pairs of pants X, Y, Z now the total no of ways of wearing the clothes = 3*3*3

Now out which XYZ, XZY, YXZ, YZX, ZXY, ZYX are favorable so the probability = 6/27 = 2/9]


Probability that he wears different shirts every day = 2/9

[ Explanation same as that of pants]


Hope it helps!
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by thephoenix » Sun Jan 31, 2010 2:34 am
its
for first day prob is
(3c1*2c1*3c1)/(3c1*2c1*3c1)

for second day prob is
(2c1*1c1*2c1)/(3c1*2c1*3c1)....rep is not allowed and for shoes its bcoz the same pair has to be selected

for third day prob is
1*1*1/(3c1*2c1*3c1)
final prob
[(3c1*2c1*3c1)*(2c1*1c1*2c1)*1*1*1]/[(3c1*2c1*3c1)*(3c1*2c1*3c1)*(3c1*2c1*3c1)]=1/3^4
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by bhumika.k.shah » Sun Jan 31, 2010 2:38 am
i like ! :-)

thank you the phoenix as usual :D
thephoenix wrote:its
for first day prob is
(3c1*2c1*3c1)/(3c1*2c1*3c1)

for second day prob is
(2c1*1c1*2c1)/(3c1*2c1*3c1)....rep is not allowed and for shoes its bcoz the same pair has to be selected

for third day prob is
1*1*1/(3c1*2c1*3c1)
final prob
[(3c1*2c1*3c1)*(2c1*1c1*2c1)*1*1*1]/[(3c1*2c1*3c1)*(3c1*2c1*3c1)*(3c1*2c1*3c1)]=1/3^4
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