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Probability Misconception

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!

Note by Sandeep Bhardwaj
6 months, 2 weeks ago

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Let \(B\) 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))\(=1-\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\boxed{\frac{7}{8}}\) Yash Dev Lamba · 6 months, 2 weeks ago

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@Yash Dev Lamba Nice approach @Yash Dev Lamba Atanu Ghosh · 6 months, 1 week ago

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@Yash Dev Lamba Yes you are right. This is the same approach what I had. Aditya Kumar · 6 months, 2 weeks ago

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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 ? Vishal Yadav · 6 months, 1 week ago

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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... Aloysius Ng · 6 months, 1 week ago

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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 ☺ Rajat Jain · 6 months, 1 week ago

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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. Alexander Koran · 6 months, 1 week ago

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How 3/4 will come? Shyambhu Mukherjee · 6 months, 2 weeks ago

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@Shyambhu Mukherjee if blindly consider BBG,BGB,GBB same and BGG,GBG,GGB also same then prob. is 3/4 which is incorrect. Yash Dev Lamba · 6 months, 1 week ago

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@Yash Dev Lamba Thanks for explaining. A wrong answer is more important than a right. Shyambhu Mukherjee · 6 months, 1 week ago

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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! Sandeep Bhardwaj · 6 months, 2 weeks ago

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@Sandeep Bhardwaj 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 ? Vishal Yadav · 6 months, 1 week ago

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Simple conceptual learning condtitonal probability Probability Rocks By- YDL Yash Dev Lamba · 6 months, 2 weeks ago

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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? Agnishom Chattopadhyay · 6 months, 2 weeks ago

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@Agnishom Chattopadhyay Ya , parodoxes created by probability are probably the best.

(a)1/3 (b)1/2 Harsh Shrivastava · 6 months, 2 weeks ago

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I feel like the answer is 1/5 Lakshya Sinha · 6 months, 2 weeks ago

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@Lakshya Sinha Can you please give some explanation supporting your answer? Sandeep Bhardwaj · 6 months, 2 weeks ago

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@Sandeep Bhardwaj 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 Lakshya Sinha · 6 months, 2 weeks ago

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@Lakshya Sinha It's only about the children, not the parents. So consider a family of 3 children (assuming total members as 3) and then find out the probability that at least one of them is a boy. Thanks!

Don't worry. Keep practicing. You will soon be a master in combinatorics. \(\ddot \smile\) Sandeep Bhardwaj · 6 months, 2 weeks ago

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@Sandeep Bhardwaj The answer is 3/4 Pawan Pal · 6 months, 2 weeks ago

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@Pawan Pal There are 4 possibilities - 1 boy , 2 boys , 3 boys and no boy . so at least 1 boy so the answer is 3/4 Am I correct ? Sandeep sir Pawan Pal · 6 months, 2 weeks ago

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@Pawan Pal No. The correct answer is 7/8. Sandeep Bhardwaj · 6 months, 2 weeks ago

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@Sandeep Bhardwaj Ohkk sir I thought order won't matter . Pawan Pal · 6 months, 2 weeks ago

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@Pawan Pal What I think is that BGG , GBG, GGB would be same so I count them as 1. Are we considering order of birth as well ? Pawan Pal · 6 months, 2 weeks ago

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@Pawan Pal If we replace chilren with coins, the answer remains the same but that maybe a better way to tell you why order is necessary. Kushagra Sahni · 6 months, 2 weeks ago

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@Pawan Pal yes, we are considering although it is not mentioned in question but it is understood (I think) to consider order of birth. Yash Dev Lamba · 6 months, 2 weeks ago

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