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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Feb 2008 15:50:45 IST
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two balls are drawn one after another(without replacement) from a bag containing 2 white,3 red ,5 blue balls .what is the probability that atleast one ball is red??
ans. 2/3
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Destiny is no matter of chance.It is a matter of choice.It is not a thing to be waited for, it is a thing to be achieved. |
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26 Feb 2008 16:09:41 IST
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i am getting 7/15??
favorable cases:
wr, br, rb, rw
wr: 2/10*3/9
br: 5/10*3/9
rw:3/10*2/9
rb:3/10*5/9
so total
= (6+15+6+15)/90
=42/90=7/15
oh sorry yeah...rr
so 3/10*2/9=1/15
so total 8/15
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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26 Feb 2008 16:12:07 IST
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im getting 8/15
you have miissed red-red....
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Destiny is no matter of chance.It is a matter of choice.It is not a thing to be waited for, it is a thing to be achieved. |
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26 Feb 2008 16:14:07 IST
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Fav. cases->wr,br,rb,rw,rr...
since 7/15 is prob. of 1st 4 cases as calc. above,1 more case is of rr whose prob. is 1/15,so total prob. is 8/15....
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26 Feb 2008 16:30:47 IST
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i think answer is printed wrong...its gotta be 8/15...we can confrm by conjugate case ttoo..
bw+wb=2*10/90=20/90
bb=5*4/90=20/90
ww=2*1/90=2/90
tot=42/90=7/15
hence conj case=1-7/15=8/15
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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26 Feb 2008 22:50:25 IST
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the ans is 8/15 as
for our requirement 2 cases arises :
1. when exactly 1 ball is red
i.e C(3,1)*C(7,1)/C(10,2)
2. when both balls are red :
i.e. C(3,2)/C(10,2)
eq1+eq2=8\15
so ans is 8\15
if ans is correct & u r satisfied with explanation do rate me
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