It's still half. Right?

Probability Level 2

In a two-child family, it is known that at least one child is a boy. What is the probability that the other child is girl?

Clarifications :

The two children are not twins.
Each child is either male or female.
Each child has the same chance of being male as of being female.
The sex of each child is independent of the sex of the other.

\frac{1}{3}

\frac{2}{3}

\frac{1}{2}

\frac{1}{4}

\frac{3}{4}

2 solutions

Yash Dev Lamba
Mar 17, 2016

This is question of conditional probabilty

Sample Space, $S=\{BB,BG,GB,GG\}$

As one child is boy, $S$ reduces to $S^{'}=\{BB,BG,GB\}$

Now, Probability that other child is girl is $= \boxed{\frac{2}{3}}.$

The answer is wrong, it should have been 1/2

Debojyoti Chakraborty - 5 years, 2 months ago

It states each baby's gender is independent of the other. How does that not imply 1/2 chance

Matthew Mostrom - 5 years, 2 months ago

BG and GB are the same thing..

T P - 5 years, 2 months ago

No , they are different because they have different relative ages (order of birth) . . Boy is elder than a girl and girl is elder than a boy are two different cases.

Yash Dev Lamba - 5 years, 2 months ago

However, per the question "The sex of each child is independent of the sex of the other." Therefore, the sample space should be S = {B, G} since one child's gender has no bearing on the other thus making the correct answer 1/2.

John Mason - 5 years, 2 months ago

I tends to saw it differently. The question does not mention the relative age anywhere, and so as the clarification. Only "not twins" are just not sufficient. I still not convincing of this answer as the sex of them should be independent.

Kay Xspre - 5 years, 2 months ago

@Kay Xspre – there is no relation each birth is 50 50

Joshua Coddington - 5 years, 2 months ago

This particular situation is a "combination" situation. Meaning order does not matter. In a "permutation" situation, order matters. So shouldn't BG and GB be the same thing? If it is infect a "permutation" situation, I'd like to know how. The situation doesn't seem to take in count the age of the children.

Adri Bhattacharjee - 5 years, 2 months ago

If you are considering that ${BG}$ is a different case from ${GB}$ you have an argumment. You say that the childs have different ages, so the cases must be different.

BUT note that if your argumment is valid, ${BB}$ should be splict into two cases: ${B1B2}$ and ${B2B1}$ , because the childs must be different (and born on distinct times). So are 4 total cases.

So $P = 2/4 = 1/2$ . Am I wrong?

Thiago Martinoni - 5 years, 2 months ago

You are right

Mohamed Refaee - 5 years, 2 months ago

If this condition - "The sex of each child is independent of the sex of the other." is included, it should be 1/2 only.

Frank Puk - 5 years, 2 months ago

I think there is a similar question in NCERT 12th standard book. Nice question though! Really helps to clear some basic concepts regarding conditional probability.

Miraj Shah - 5 years, 2 months ago

This is a fairly controversial question which the mathematics community has debated for decades; http://mathforum.org/library/drmath/view/52186.html

John Mason - 5 years, 2 months ago

is this a correct solution? one child is surely boy. so the other child could be boy or girl. {B,G}; so the probability should be 1/2

Asma Ul Chowdhury - 5 years, 2 months ago

https://en.wikipedia.org/wiki/Boy or Girl paradox#Second question

Bob Bob - 5 years, 2 months ago

The answer is wrong, because, as the gender of the first child is already known, we are left with only two equally likely choices for the second child. Thus, the probability of the second child being a girl is 1/2. Had the question been like this: "What is the probability that at least one of two children in a family is a girl?" Then the answer will definitely be 2/3.

Solomon Hailu - 5 years, 2 months ago

Lokesh Sharma
Apr 14, 2016

It's a trick question. I have a simulation in Ruby below that proves that the answer should be 2/3.

# Randomly have a bor or a girl
def get_a_child
  [:boy, :girl].sample
end

# Create families under the assumptions in the problem
def create_families number_of_families
  number_of_families.times.map { [get_a_child, get_a_child] }
end

# Let's see how many of 1000 families have atleast one boy
all_familties = create_families 10000
atleast_one_boy = all_familties.select { |f| f.include? :boy }

# Let's see how many of above chosen families have other child as a girl.
another_child_girl = atleast_one_boy.select { |f| f.include? :girl }

puts "Out of 10000"
puts "x = Families with atleast one boy: #{x = atleast_one_boy.size}"
puts "y = Above families having other child as girl: \
#{y = another_child_girl.size}"
puts "y/x = #{y/x.to_f}"

Output:

Out of 10000
x = Families with atleast one boy: 7487
y = Above families having other child as girl: 5034
y/x = 0.6723654334179244

The chances of any one child being a girl is 50% - the other siblings don't come into account.

Holly McNeill - 5 years, 2 months ago

Hard to debate on that. All I can say is I played by the rules and there's the result. I didn't assumed any probability concept while calculating the probability. If you think there's problem with the way I've simulated the problem let me know.

Lokesh Sharma - 5 years, 2 months ago

It's still half. Right?

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