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

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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