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by sanju09 » Tue Mar 02, 2010 4:22 am
10 different books and 2 different pens are given to 3 boys so that each boy gets equal number of things. What is the probability that same boy doesn't receive both pens?
(A) 3/11
(B) ¼
(C) 4/11
(D) 8/11
(E) 9/11
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by rohan_vus » Tue Mar 02, 2010 4:38 am
Probability that one boy receives both pens is 10c2*2c2/12c4 = 1/11
So total probability that among 3 boys, anyone will have boths pens, is 3*1/11 = 3/11

So probability that none will have both pens = 1 - 3/11 = 8/11
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by stufigol » Wed Mar 03, 2010 12:13 am
u explain the 10c2 and 12c4 plz
thanks
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by harsh.champ » Wed Mar 03, 2010 4:27 am
stufigol wrote:u explain the 10c2 and 12c4 plz
thanks
Ok.
Lets look at rohan's explanation.
Probability that one boy receives both pens is 10c2*2c2/12c4 = 1/11
So total probability that among 3 boys, anyone will have boths pens, is 3*1/11 = 3/11

So probability that none will have both pens = 1 - 3/11 = 8/11
Suppose the same boy gets both the pens.
Now,10C2 = No. of ways in which the 2 books can be selected.
12C4 = Selection of any 4 items out of the (10 + 2) items

Also,the boy can be selected in 3C1 ways.
So, we have our case (10C2 x 2C2 x 3C1)/12c4 = 3/11
Now, P(favorable case) = 1 - P(all the unfavorable cases)

I hope it is clear now. :)
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