# Probability: Consider a family with 5 children...

If it is known that the family has at least one boy, what is the probability that the oldest child is a boy?

My attempt at this: The probability that the known boy, call it $boy_0$ is the oldest is:

$p(boy_0)= \frac{1}{5}$

Among the other children the probability of being a boy and the oldest is:
$p(boy_{old}): \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10}$

Now, the probability of being a boy and the oldest for the known and unknown sets of children is

$p(boy_0)+p(boy_{old})\frac{1}{5} + \frac{1}{10} =\frac{3}{10}$

Am I correct here?

What is the probability that there are 2 boys and 3 girls?

My attempt:

Probability that there are 2 boys out of 5 children is:

$p(boy) = {5\choose 2} (0.5)^2 (1-0.5)^{5-2}= \frac{5}{16}$

Probability that there are 3 girls out of 5 children is:

$p(girl)= {5\choose 3} (0.5)^3 (1-0.5)^{5-3} = \frac{5}{16}$

Probability that there are 2 boys and 3 girls is: $p(boy and girl) = p(boy) \cdot p(girl) = \frac{5}{16} \cdot \frac {5}{16} = \frac{25}{256}$ Correct or not?

Lastly,

In this family of 2 boys and 3 girls, what is the probability that the oldest child is a boy?

Probability of being oldest is $\frac{1}{5}$ and probability of being a boy is$\frac{2}{5}$. Thus the probability of the oldest being a boy is $\frac{1}{5} \cdot \frac{2}{5} = \frac{2}{10}.$

I'm a bit shakey about all my answers. Are any of them correct?

For part I:

Order the children from youngest to oldest. A priori there are $2^5=32$ equally probable ways to assign gender to the collection. We exclude "all girls", making for $31$ cases. Other than $\{G,G,G,G,B\}$ (a $\frac 1{31}$ case), every combination with a boy in the last spot can be paired with one with a girl in the last slot. Thus the probability that the eldest is a boy is $$\frac 12\times \frac {30}{31}+\frac 1{31}=\frac {16}{31}$$

For part II:

your first line already specifies two boys and three girls, you don't need the second factor.

For part III:

There are $10$ equally probable scenarios in which there are $2$ boys and $3$ girls. $4$ of these have the eldest child a boy, so $\frac 4{10}$.

• Hi, how did you figure out the numbers for parts 1 and 3? Where did 30/31 originate and how did you calculate 4/10? Nov 4, 2016 at 0:06
• For part 1: excluding the case I mentioned there are $15$ scenarios with an eldest boy and $15$ with an eldest girl, making for $\frac {15}{31}$ in probability. Then you need to add back the isolated case. For part 3: there are $10$ admissible scenarios, $4$ of which have an eldest boy.
– lulu
Nov 4, 2016 at 0:10
• I suggest: Work the same problem (part I) with $3$ kids. Now there are only $8$ starting scenarios, so you can just write them all down. My formula yields $\frac 12\times \frac 67+\frac 17=\frac 47$. For that matter, it's not all that hard to write out all $32$ cases in the actual problem.
– lulu
Nov 4, 2016 at 0:11
• Ah, I see, thank you very much. Nov 4, 2016 at 0:16

The probability that a family of five has at least one boy is: the probability that they do not have all girls: $$1-(\tfrac 12)^5 = \tfrac{31}{32}$$

The probability that a family of five has a boy as the eldest child, is: $$(\tfrac 12) = \tfrac {16}{32}$$

Apply the very definition of a modern major general conditional probability:$$\dfrac{\tfrac 12}{1-{(\tfrac 12)}^5} = \dfrac{2^4}{2^5-1}$$

Therefore the probability that a family of five has a boy for the eldest child when given that they have at least one boy is: $16/31$

Likewise the second part: In a family of five with at least one boy, the probability that there are exactly two boys is

$$\dfrac{\binom 52 2^{-5}}{1-2^{-5}} = \dfrac{10}{31}$$

And when given that there are 2 boys and 3 girls the probability that the eldest is a boy is, indeed, $2/5$.