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!

#Combinatorics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 3 months ago

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

Aditya Kumar - 5 years, 3 months ago

Nice approach @Yash Dev Lamba

Atanu Ghosh - 5 years, 2 months ago

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

A Former Brilliant Member - 4 years, 3 months ago

lol.

Swapnil Das - 2 years, 8 months ago

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 - 5 years, 3 months ago

Ya , parodoxes created by probability are probably the best.

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

Harsh Shrivastava - 5 years, 3 months ago

Simple conceptual learning condtitonal probability Probability Rocks By- YDL

Yash Dev Lamba - 5 years, 3 months ago

@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 - 5 years, 2 months ago

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 - 5 years, 2 months ago

How 3/4 will come?

Shyambhu Mukherjee - 5 years, 2 months ago

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

Yash Dev Lamba - 5 years, 2 months ago

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

Shyambhu Mukherjee - 5 years, 2 months ago

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 - 5 years, 2 months ago

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 - 5 years, 2 months ago

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 - 5 years, 2 months ago

Vishal Yadav - 5 years, 2 months ago

I feel like the answer is 1/5

Department 8 - 5 years, 3 months ago

Can you please give some explanation supporting your answer?

Sandeep Bhardwaj - 5 years, 3 months ago

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

Department 8 - 5 years, 3 months ago

@Department 8 – 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 - 5 years, 3 months ago

@Sandeep Bhardwaj – The answer is 3/4

Pawan pal - 5 years, 3 months ago

@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 - 5 years, 3 months ago

@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 - 5 years, 3 months ago

@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 - 5 years, 2 months ago

@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 - 5 years, 3 months ago

@Pawan Pal – No. The correct answer is 7/8.

Sandeep Bhardwaj - 5 years, 2 months ago

@Sandeep Bhardwaj – Ohkk sir I thought order won't matter .

Pawan pal - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$