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# A couple has 3 children. Find the probability that atleast

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Manager
Joined: 15 Jul 2005
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A couple has 3 children. Find the probability that atleast [#permalink]

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06 Oct 2005, 22:06
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A couple has 3 children. Find the probability that atleast one is a boy. Assume likelyhood of having a boy or a girl is equal and independent.

a) 1/8
b) 1/3
c) 2/7
d) 7/8
e) 5/8

Last edited by chets on 06 Oct 2005, 22:17, edited 1 time in total.

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GMAT Club Legend
Joined: 07 Jul 2004
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06 Oct 2005, 22:12
P(Boy) = P(Girl) = 1/2
Events are independent, mutually exclusive.

P(one boy) = P(Boy and Girl and Girl) = 1/2 * 1/2 * 1/2 = 1/8

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Manager
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06 Oct 2005, 22:17
Sorry, I made a mistake in the question.

It should read, find the probability that atleast 1 is boy.

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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5032

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06 Oct 2005, 22:24
chets wrote:
Sorry, I made a mistake in the question.

It should read, find the probability that atleast 1 is boy.

Then probability = 1 - P(all girls) = 1 - 1/8 = 7/8

Ans: D

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SVP
Joined: 05 Apr 2005
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06 Oct 2005, 22:48
P(0 boy) = (1/2)^3=1/8
so, P(at least a boy)=total prob-P(0 boy)=1-1/8=7/8

alt: p(a boy)=1/2, p(2 boy)=1/4, p(3 boy)=1/8
so, P(at least a boy) = p(a boy)+p(2 boy)+p(3 boy)=1/2+1/4+1/8= 7/8

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Senior Manager
Joined: 15 Apr 2005
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06 Oct 2005, 22:59
chets wrote:
A couple has 3 children. Find the probability that atleast one is a boy. Assume likelyhood of having a boy or a girl is equal and independent.

a) 1/8
b) 1/3
c) 2/7
d) 7/8
e) 5/8

P(Atleast 1 boy) = 1 - P(All being girls)
= 1 - (1/2*1/2*1/2)
= 1 - 1/8
= 7/8

Kudos [?]: 17 [0], given: 0

Senior Manager
Joined: 22 Jun 2005
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07 Oct 2005, 02:20
It's D

The probability of not having any boy is
1/2*1/2*1/2=1/8

The probability of having at least one boy:

1-1/8 = 7/8.

Kudos [?]: 11 [0], given: 0

07 Oct 2005, 02:20
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