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Updated on: 13 Mar 2019, 05:07
Originally posted by petrifiedbutstanding on 06 Feb 2011, 02:20.
Last edited by Bunuel on 13 Mar 2019, 05:07, edited 1 time in total.
Last edited by Bunuel on 13 Mar 2019, 05:07, edited 1 time in total.
Renamed the topic and edited the question.
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D
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Difficulty:
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Question Stats:
65% (01:06) correct
35%
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based on 576
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What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 5/8
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 5/8
06 Feb 2011, 02:52
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Probability(1B,2G) = Favorable choices of having exactly 1B and 2G/Total number of Possiblities of sexes for 3 children
Let's do the sample set:
Represent Girls with G and Boys with B
3 children can have following possibilities of sexes
GGG - 3 girls
GGB - 2 girls one boy
GBG
GBB
BGG
BGB
BBG
BBB
Total possibilities of both sexes = 8
Number of choices where there are exactly 2 girls and 1 boy are:
GGB
GBG
BGG
=3.
Thus, probability = 3/8
Ans: "D"
This problem is similar to having exactly 2 heads in 3 tosses.
Alternate way;
The total number of possibilities = (Number of possible outcomes in each flip)^(Number of tosses) = 2^3=8
Likewise;
The total number possible sexes for 3 children = (Number of possible sex for each child)^(Number of children)
Number of possible sex for each child = 2 = (Boy or Girl)
Number of children = 3
Total possible sexes = 2^3 = 8
Possibilities to have exactly 2 Girls out of 3 child and 1 Boy out of remaining 1 Child:
=
\(C^{3}_{2}*C^{1}{1} = 3\)
Thus, probability = Favorable/total outcomes = 3/8.
Let's do the sample set:
Represent Girls with G and Boys with B
3 children can have following possibilities of sexes
GGG - 3 girls
GGB - 2 girls one boy
GBG
GBB
BGG
BGB
BBG
BBB
Total possibilities of both sexes = 8
Number of choices where there are exactly 2 girls and 1 boy are:
GGB
GBG
BGG
=3.
Thus, probability = 3/8
Ans: "D"
This problem is similar to having exactly 2 heads in 3 tosses.
Alternate way;
The total number of possibilities = (Number of possible outcomes in each flip)^(Number of tosses) = 2^3=8
Likewise;
The total number possible sexes for 3 children = (Number of possible sex for each child)^(Number of children)
Number of possible sex for each child = 2 = (Boy or Girl)
Number of children = 3
Total possible sexes = 2^3 = 8
Possibilities to have exactly 2 Girls out of 3 child and 1 Boy out of remaining 1 Child:
=
\(C^{3}_{2}*C^{1}{1} = 3\)
Thus, probability = Favorable/total outcomes = 3/8.
General Discussion
06 Feb 2011, 02:59
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fluke
Thanks. This explanation is much simpler than the one in the source. I've been using the F/T rule in most of my previous problems.
But the explanation in the source confused me.
