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Probability of a couple having a boy child is 0.4. If a couple decides
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21 Aug 2023, 03:49
21 Aug 2023, 03:49
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Probability of a couple having a boy child is 0.4. If a couple decides to have 3 children, what is the probability that atleast one of them is a girl child?
A. 0.064
B. 0.36
C. 0.64
D. 0.936
E. 1
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A. 0.064
B. 0.36
C. 0.64
D. 0.936
E. 1
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
\(\Longrightarrow\) ALL the GRE Quant and Verbal Practice Directories in ONE place (2023)
\(\Longrightarrow\) Best GRE Test Books of 2023
Re: Probability of a couple having a boy child is 0.4. If a couple decides
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21 Aug 2023, 04:11
21 Aug 2023, 04:11
1
We are told to find the probability that at least one of the children is a female.
Since these are independent think (similar to the case of a coin flip, heads or tails) we can create two "sets" of events:
P(all boys)+P(at least 1 girl)=1
P(at least 1 girl)=1-p(all boys)
This will shorten the time on the problem considerably - there is only 1 possible combination for all boys rather than the 3 combos for at least 1 girl.
P(all boys)=(2/5)^3=8/125
1-(8/125)=117/125=0.936
Choice D is correct
Since these are independent think (similar to the case of a coin flip, heads or tails) we can create two "sets" of events:
P(all boys)+P(at least 1 girl)=1
P(at least 1 girl)=1-p(all boys)
This will shorten the time on the problem considerably - there is only 1 possible combination for all boys rather than the 3 combos for at least 1 girl.
P(all boys)=(2/5)^3=8/125
1-(8/125)=117/125=0.936
Choice D is correct
Re: Probability of a couple having a boy child is 0.4. If a couple decides
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20 Dec 2023, 22:37
20 Dec 2023, 22:37
if i go with complementary method : 1 - P(BBB) = 1- (0.4×0.4×0.4) = 0.936
but Carcass sir, why cant we find the answer through below approach , what is flaw here ?
P(at least one girl)=P(GBB)+P(GGB)+P(GGG)
= (0.6×0.4×0.4)+(0.6×0.6×0.4)+(0.6×0.6×0.6)
= 0.456
but Carcass sir, why cant we find the answer through below approach , what is flaw here ?
P(at least one girl)=P(GBB)+P(GGB)+P(GGG)
= (0.6×0.4×0.4)+(0.6×0.6×0.4)+(0.6×0.6×0.6)
= 0.456
