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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Feb 2008 19:27:54 IST
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PLz help me solve these probabiliy problems.I tried but my answers weren't matching with tht given in the book.below are answers from the book : !.)A committe of 4 boys and 3 girls is to be formed from a group of 8 boys n 5 girls selecting randomly.What is the probability that the committe contains a particular boy n a particular girl?Ans..3/10 2.)A card is drawn from a pack of 52 cards.Find the probability that the card drawn is red or bears a number between 5 n 10 both exclusive.Ans..19/26 3.)two cards are drawn at random from 52 playing cards.Find the probability tht one is a spade and the other is an aceAns..1/34
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12 Feb 2008 19:35:47 IST
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plz it's urgent
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12 Feb 2008 19:53:13 IST
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plz don't anyone know how to solve it plz it's urgent
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12 Feb 2008 19:53:37 IST
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a particular boy n a particular girl has to be selected
hence remaining 3 boys can be selected in 7c3 ways
similarly remaining 2 girls be selected in 4c2 ways
hence possible ways =7c3* 4c2
=7*6*5/3*2*1 *4*3/2*1 =210
total way =8c4* 5c3
= 700
p=210/700=3/10
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12 Feb 2008 19:58:12 IST
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n(s)=52c1=52
n(a)=26+24-12=38 { n(a u b)= n(a) + n(b) -n(a intersection b)}
p=38/52=19/26
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12 Feb 2008 20:00:06 IST
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thnx my mistake was tht i took sample space as 13C7.I added the girls n boys .thnx
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12 Feb 2008 20:04:05 IST
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but for Q.2) 5 n 10 both are exclusive so thr will be 4*4=16 cards with numbers 6,7,8,9
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12 Feb 2008 20:13:14 IST
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did u get it.plz reply don't worry about the rates
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12 Feb 2008 20:23:30 IST
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don't anyone want to answer ,rates assured
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13 Feb 2008 18:03:54 IST
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26+16-8=34
p=34/52=17/26
does not match with ur answer
seems either it is ur mistake or printing mistake
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13 Feb 2008 18:06:58 IST
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3
n(s)= 52c2=51*52/2
n(a)=13c1 +4c1 -1=13+4-1=16
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14 Feb 2008 00:09:47 IST
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3. probability = (4c1/52c1) +2c1/52c1 - 1/52c1
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