Hey, buddies :)

Recently people had discussion in Brilliant-Lounge on a probability problem which is:

In a family of 3 children, what is the probability that at least one will be a boy?

Some of them believe that \(\frac 34\) is the correct answer while the others believe that the correct answer is \(\frac 78\).

Everyone is invited to come up with their response along with the explanation. It will be fun and help us a lot to upgrade our knowledge engine further.

Thanks!

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## Comments

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TopNewestLet \(B\) represent a boy and \(G\) represent a girl.Then,

Sample Space , \(S =\{BBB,BBG,BGB,GBB,BGG,GBG,GGB,GGG\}\)

(Probability of having at least \(1\) boy) \(= 1 - \)(Probability of having only girls (no boys))=\(1-\frac{1}{8}\)(only \(1\) case out of \(8\) cases)=\(\frac{7}{8}\)

So According to me, correct answer is \(\boxed {\frac{7}{8}}\).

Note:-\(BBG,BGB,GBB\) are different cases because their relative ages (order of birth) are different in each case.

(Same for \(BGG,GBG,GGB\))

Alternate Thinking Process:-\(P(B)=P(G)=\frac{1}{2}\)

(Probability of having at least \(1\) boy) \(= 1 - \)(Probability of having only girls (no boys))\(=1-\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\boxed{\frac{7}{8}}\)

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Nice approach @Yash Dev Lamba

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Yes you are right. This is the same approach what I had.

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it's an easy one :P let the birth of boy be success and girl be failure ( i'm not being an anti-feminist :P) the answer would be 3c1+3c2+3c3/(2^3)=7/8 ! easy enough :P

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lol.

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The Family cares about getting Boy, not about getting a young boy or an old boy. So, how does having two younger daughters and an elder son different from having a younger son and two elder daughters ?

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But there is a large assumption that boys and girls are given birth to with 0.5 probability each! That's quite huge an assumption, and it would allow any answer to be correct...

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Ans should be 7/8 as if we remove the case of no boys then the remaining will be atleast 1 boy i.e 1-(1/2)^3= 7/8 ☺

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Take the complementary probability, the chance that no boys are picked. For this to happen, all girls must be picked, so the probability is (1/2)^3 = 1/8. Every other case has at least one boy, so the probability that at least one boy is chosen is 1 - 1/8 = 7/8.

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How 3/4 will come?

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if blindly consider BBG,BGB,GBB same and BGG,GBG,GGB also same then prob. is 3/4 which is incorrect.

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Thanks for explaining. A wrong answer is more important than a right.

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@Nihar Mahajan @Sharky Kesa

What do you guys think? I would be great if you participate in this discussion as you were playing a major role in the slack discussion. Thanks!

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The Family cares about getting Boy, not about getting a young boy or an old boy. So, how does having two younger daughters and an elder son different from having a younger son and two elder daughters ?

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Simple conceptual learning condtitonal probability Probability Rocks By- YDL

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I agree with @Yash Dev Lamba

Probability can lead to amazing paradoxes. Here is a very well known probability question often misunderstood:

A family has two children. What is the probability that they are both sons, given that

a)At least one of them is a son?b)the elder child is a son?Log in to reply

Ya , parodoxes created by probability are probably the best.

(a)1/3 (b)1/2

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I feel like the answer is 1/5

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Can you please give some explanation supporting your answer?

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Duh!! It is 4/5 because in a family of 5 (3 children 2 parents) then there is at least 1 women so at least 1 boy is 4/5

PS: I am weak in probability

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Don't worry. Keep practicing. You will soon be a master in combinatorics. \(\ddot \smile\)

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