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Three brothers apply to a certain business school. The

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Three brothers apply to a certain business school. The [#permalink] New post 29 Jun 2004, 06:48
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Three brothers apply to a certain business school. The probability of the first one being admitted is 0.9; the second—0.5; the third—0.2. Find the probability that:

1) at least two are admitted
2) at least one is admitted
3) none is admitted
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 [#permalink] New post 29 Jun 2004, 06:58
1) Atleast two --> 0.9*0.5+0.5*0.2+0.9*0.2+0.9*0.5*0.2 = 0.82
2) Atleast One --> 1 - None --> 1 - (1-0.9)*(1-0.5)(1-0.2) = 1 - 0.1*0.5*0.8 = 0.96
3) None (1-0.9)*(1-0.5)(1-0.2) = 0.04
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 [#permalink] New post 29 Jun 2004, 07:43
Ya! sure about all three.

Why? is the OA different??
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Re: PS: three brothers--probability [#permalink] New post 29 Jun 2004, 09:31
stolyar wrote:
Three brothers apply to a certain business school. The probability of the first one being admitted is 0.9; the second—0.5; the third—0.2. Find the probability that:

1) at least two are admitted
2) at least one is admitted
3) none is admitted


1) P(2 get in) + P(3 get in)=
.9(.5)*(1-.2)=.36
.9(.2)*(1-.5)=.09
.5(.2)*(1-.9)=.01
.36+.09+.01=.46 -->P2 get in
(mental check, 3C2=3 different ways to get 2 out of three)

P(3 get in) = .9(.5)(.2)=.09

P(2 get in) + P(3 get in) = .55

(from a quick glance, I think MBA forgot to multiply the probabilities of 2 get in by the probability of the 3rd person not getting in, hence his answer was of)

2) P (at least one gets in) = 1 - P(non are admitted)
1- (1-.9)(1-.5)(1-.2) = 1-.04=.96

3)P (none get in) = .04
Re: PS: three brothers--probability   [#permalink] 29 Jun 2004, 09:31
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