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# There are 18 shirts in a closet. 12 are stripes and 6 are

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There are 18 shirts in a closet. 12 are stripes and 6 are [#permalink]

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31 Mar 2008, 16:55
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There are 18 shirts in a closet. 12 are stripes and 6 are white. 6 draws are made of 1 shirt at a time without any replacement of which at least 4 are found to be white. What is the probability that in the next 2 draws, exactly 1 shirt is white?
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31 Mar 2008, 17:11
After 4 whites are gone, the there are only 2 white shirts left in the closet.
So number of ways 1 shirt can be chosen from 2 shirts = 2C1 = 2

There are total of 12 stripes.
So number of ways 1 stripes shirt can be chosen from 12 shirts = 12C1 = 12

Total number of shirts in closet = 2 + 12 = 14
Total number of ways of selecting 2 shirts from 14 = 14C2 = 91

So probability = 2*12/91 = 24/91

However please note that white shirt can be chosen first and stripes second and vice-versa is also possible.
So final probability needs to be multiplied by 2!.

Probability = 24*2!/19 = 48/91
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31 Mar 2008, 17:39
Can you explain this part of your answer? I did not understand this. I got till 14C2 = 91

So probability = 2*12/91 = 24/91

However please note that white shirt can be chosen first and stripes second and vice-versa is also possible.
So final probability needs to be multiplied by 2!.

Probability = 24*2!/19 = 48/91
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31 Mar 2008, 17:48
Question tells us that out of remaining 2 shirts 1 has to be white and other 1 has to be stripes. But question does not tells you the order in which they will be drawn.
So for 1 case, white can be first drawn and stripes can be drawn later.
But it is also possible that stripes is drawn first and then white is drawn later.
Thereby there are 2 ways by which we can draw one white and one stripes from the closet.
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31 Mar 2008, 17:51
Thank you
Current Student
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01 Apr 2008, 09:38
sorry, i don't have the OA for this one.
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01 Apr 2008, 11:58
Expert's post
young_gun wrote:
There are 18 shirts in a closet. 12 are stripes and 6 are white. 6 draws are made of 1 shirt at a time without any replacement of which at least 4 are found to be white. What is the probability that in the next 2 draws, exactly 1 shirt is white?

First of all, I would say that this problem is not like GMAT one.

Here, we have two events:

1. 6 draws are made of 1 shirt at a time without any replacement of which at least 4 are found to be white.
So, we have to consider 3 possibilities: 4 white and 2 stripes shirts, 5 white and 1 stripes shirts, and 6 white and 0 stripes shirts:

a) 4 white and 2 stripes shirts: the number of combination: $$N_a=C^6_4*C^{12}_2=\frac{6*5}{2}*\frac{12*11}{2}=990$$

b) 5 white and 1 stripes shirts: the number of combination: $$N_b=C^6_5*C^{12}_1=6*12=72$$

c) 6 white and 0 stripes shirts: the number of combination: $$N_c=C^6_6*C^{12}_0=1*1=1$$

Therefore, the probabilities will be:

a) $$p_a=\frac{990}{990+72+1}=\frac{990}{1063}$$

b) $$p_b=\frac{72}{1063}$$

c) $$p_c=\frac{1}{1063}$$

2. in the next 2 draws, exactly 1 shirt is white

a) $$q_{a}=\frac{990}{1063}*\frac{C^2_1*C^{10}_1}{C^{14}_2}=\frac{990}{1063}*\frac{40}{182}$$

b) $$q_{b}=\frac{72}{1063}*\frac{C^1_1*C^{11}_1}{C^{14}_2}=\frac{72}{1063}*\frac{22}{182}$$

c) $$q_{c}=\frac{72}{1063}*0=0$$

Finally,

$$P=q_a+q_b+q_c =\frac{990}{1063}*\frac{40}{182}+\frac{72}{1063}*\frac{22}{182}+0 =\frac{1062}{1063}*\frac{62}{182} = \frac{1062}{1063}*\frac{31}{91}$$

I understand that I am wrong but do not understand where I am wrong
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01 Apr 2008, 17:38
walker wrote:
young_gun wrote:
There are 18 shirts in a closet. 12 are stripes and 6 are white. 6 draws are made of 1 shirt at a time without any replacement of which at least 4 are found to be white. What is the probability that in the next 2 draws, exactly 1 shirt is white?

I understand that I am wrong but do not understand where I am wrong

First of all question is saying 4 shirts are already drawn they are all white. So that part is already out of scope, no need to factor that calculating final probability. Moreover question is saying that in remaining 2 shirts exactly 1 of them should be white, so cases of 6 white and 4 white and 2 stripes become invalid and should not be factored for final calculations.
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01 Apr 2008, 23:30
young_gun wrote:
There are 18 shirts in a closet. 12 are stripes and 6 are white. 6 draws are made of 1 shirt at a time without any replacement of which at least 4 are found to be white. What is the probability that in the next 2 draws, exactly 1 shirt is white?

My take

2 things to note:
A. without any replacement
B.at least 4 are found to be white

A =>
6 shirts are already out so total number of shirts is 12 for next 2 draws.

B =>
we have 3 cases
Case 1)4 white shirts and 2 striped shirts taken out => 2 white & 10 striped remain
Case 2)5 white shirts and 1 striped shirt taken out => 1 white and 11 striped remain
Case 3)6 white shirts and 0 striped shirt takes out => 0 white and 12 striped remain

What we want is
>the probability that in the next 2 draws, exactly 1 shirt is white?

for case 1
P(W|S) = ( 2/12 * 10/11) / 10/11 OR
P(S|W) = (10/12 * 2/11)/ 2/11

P(case 1) = 1 (????)

for case 2

P(W|S) = ( 1/12 * 11/11) / 11/11 OR
P(S|W) = (11/12 * 1/11)/ 1/11

so case 3) is ignored as no white shirts present. = 0

&.... I have screwed somewhere!
Re: PS probability--shirts   [#permalink] 01 Apr 2008, 23:30
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