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

Question

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?

itishaj · Answer

31/132

first case: in the first draw..5 white shirts were picked and one striped.
nw proabilty of picking exactly one shirt in the next draw is 1/12 * 11/11
i.e 1/12

second case: 4 white shirts and 2 striped shirts in d first draw
nw proabilty of picking exactly one shirt in the next draw is 2/12 * 10/11
i.e 5/33

total prob is 1/12 + 5/33 = 31/132

Professor · Answer

this is conditional prob, i am not sure with my solution.. 

*suppose we already have 4 whites, now the possibilities of getting 1 white (it could be 5th drawing or 6th drawing) = 2(10/12c2) = 20/33
*suppose we already have 5 whites, now the possibilities of getting 1 white (it could be 5th drawing or 6th drawing) = 2(11/11c2) = 22/55 = 2/5
*suppose we already have 6 whites, now the possibilities of getting 1 white = 0

so the prob = 20/33 + 2/5

itishaj · Answer

Agree with professor's point that we have to consider both 5 th and 6th draw.
Howver, if I modify my solution(given above) to incorporate this condition..
I still get ...
first case: in the first draw..5 white shirts were picked and one striped. 
nw proabilty of picking exactly one shirt in the next draw is 1/12 * 11/11 
+ 11/12 * 1/11 = 1/6

Professor , I think in ur solution....it should be 12C2...(marked in red)
 
second case: 4 white shirts and 2 striped shirts in d first draw 
nw proabilty of picking exactly one shirt in the next draw is 2/12 * 10/11 
+ 10/12 * 2/11 = 10/33

however, prof's answer for this case is 20/33...can smbdy clear ot the discrepancy...

total prob is 1/6 + 10/33 = 31/66

ywilfred · Answer

18 shirts: 12 stripes, 6 white

6 draws made, at least 4 white

So we could have 4 white shirts, 5 white shirts, 6 white shirts. Consider case by case:

6 white shirts: P(next two draw at least 1 is white) = 0

5 white shirts: Can be 7th draw = white or 8th draw = white
P = (1/12)(1) + (11/12)(1/11) = 1/6

4 white shirts: Can be 7th draw = white, or 8th draw = white
P = (2/12)(10/11) + (10/12)(2/11) = 10/33

Total = 31/66

ksb · Answer

At least 4 out of the 6 draws is white means 
 4 are white or 5 are white or all 6 are white.

Case 1: exactly 4 are white in the first 6 draws. Therefore, the other 2 are stripes. 
The probability of getting exactly 4 white and 2 stripes from 6 whites and 12 stripes is (6C4 * 12C2)/18C6.
Now, getting excatly one of the other two draws i.e., 7th or 8th as white from the remaining 2 whites and the other one stripe from 10 stripes is
2C1*10C1/12C2.
Combining the two, we get 6C4 * 12C2 *2C1 * 10C1/(12C2 * 18C6)
= 6C4 * 2C1 * 10C1 / 18C6

Case 2: exactly 5 are white in the first 6 draws. Therefore, the other one is a stripe
probability of getting 5 whites from 6 and 1 stripe from 12 = 6C5 * 12C1/18C6

Getting one of 7th or 8th as stripe and the other as white is 1C1 * 11C1/12C2

Combining the two, we get 6C5 *12C1 * 11C1/(18C6 * 12C2)

If you get 6 whites in the first 6 draws as professor had mentioned, you will not be able to get exactly 1 white in the next two draws.

Therefore, final probability is {6C4 * 2C1 * 10C1 / 18C6 } + {6C5 *12C1 * 11C1/(18C6 * 12C2)}

remgeo · Answer

Agree with 31/66. 
Same method as ywilfred

Professor · Answer

oh, i agree that it should be 12c2. thanxxxx.

i am doing it again.

*suppose we already have 4 whites, now the possibilities of getting 1 white (it could be 5th drawing or 6th drawing) = 2(10/12c2) = 10/33

*suppose we already have 5 whites, now the possibilities of getting 1 white (it could be 5th drawing or 6th drawing) = 2(11/12c2) = 1/3 
 . 

*suppose we already have 6 whites, now the possibilities of getting 1 white = 0

so the prob = 10/33 + 1/3 = 21/33

saha · Answer

We are given at least 4 whites but the probability of 4W2S, 5W1S and 6W are not the same so we should take the conditional probablity into account.

P(4w2S/4W) = 6c4 * 12c2/(6c4*12c2+6c5*12C1 + 6c6)
P(5w1s/4w) = 6c5 * 12c1/(same denom as above)
P(6w/4w) = 6c6 /(same denom as above)

we can now plug in each of the 3 values above.

P = P(1w/4w2s) * P(4w2s/4w) + P(1w/5w1s)*P(5w1s/4w) + P(1w/6w)*P(6w/4w)

Plug in the terms from the quoted post and above for the answer. Too tedious for me.

old_dream_1976 · Answer

If 5of6 shirts are white =   +   
 7th is w 8th is NW 7th is NW 8th is W

if 4 of 6 are wihite =   +   +  


=  / = 16/33

OA please

x242222 · Answer

I got 62/132 which is 31/66.

tl372 · Answer

I did it the same way at itishaj and got 31/66.

old_dream_1976 · Answer

nevermind - Please ignore this post

treefit · Answer

what does the "C" in the notation "12C2" stand for?

mahesh004 · Answer

I totally agree with ywilfred's method...

good problem.

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

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