Probability - Interview Question Hi I recently got this question in an interview and would like help solving it. 
There were 2 boys and 3 girls in a room. A new baby entered the
room whose gender was unknown (but equally likely to be a boy/girl).
A nurse came into the room and picked one baby randomly and it
turned out to be a boy. What’s the probability that the new baby was
a boy?
 A: Let $bg$ denote the event that the new child is a boy, and the randomly chosen child is then a girl. Label other events similarly viz. $P(bb)=P(bg)=\frac{1}{4},\,P(gb)=\frac{1}{6},\,P(gg)=\frac{1}{3}$. Then $$P(\text{new boy}|\text{boy chosen})=\frac{P(bb)}{P(bb)+P(gb)}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{6}}=\frac{3}{5},$$where to simplify I've multiplied numerator and denominator by $12$.
As an existing answer that reached this value has been downvoted, I double-checked my answer with the Python here. In one run, the conditional probability estimate was $0.602380952381$, which is close enough to $0.6$ for the purposes of a sanity check.
A: I won't spoil the answer for you, but think of what happens(intuitively) in extreme cases:
Suppose there were 1000 girls in that room and 1 boy. The nurse goes in and picks up a boy. What's the probability that the newborn was a boy?
Suppose there were 1000 boys and 1 girl. The nurse goes in and picks up a boy. What's the probability the newborn was a boy?
Use conditional probability to formalize your intuition.
A: The answer is $\boxed{3/5}$
Bayes' Rule asserts that the conditional probability of event $A$ happening given that event $B$ occurred is given by 
$$P(A\mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$
Here, your event $A$ is the event that the new child is a boy. Your event $B$ is the event of selecting one child and it being a boy. 
$P(A) = 1/2$ as you mentioned.
To compute $P(B)$, note that there's a $1/2$ chance of having a boy. Moreover, if the new child is a boy, the probability of choosing a boy becomes $1/2$. Now, the probability that both of these events occur is given by 
$$1/2 \cdot 1/2 = 1/4.$$
Now, consider the case where the child is a girl. This happens again with probability one half, and if it is a girl the probability of choosing a boy becomes $2/6$. Putting these together, we get 
$$1/2 \cdot 2/6= 1/6.$$
So, $P(B) = 1/4 + 1/6= 5/12$.
Finally, $P(B \mid A)$ is the chance of choosing a boy given that the newborn is a boy, which occurs with probability $1/2$. Plugging these into our equation provided by Bayes' Rule, we get $P(A \mid B) = 3/5.$
A: $p(new\ baby=boy|picked\ baby=boy)=\frac{p(new\ baby=boy\ and \ picked\ baby=boy)}{p(picked\ baby=boy | new\ baby=boy)p(new\ baby=boy)+p(picked\ baby=boy | new\ baby=girl)p(new\ baby=girl)}$
$p(new\ baby=boy|picked\ baby=boy)=\frac{1/2x1/2}{1/2x1/2+1/3x1/2}=3/5$
A: Note that first a new baby entered the room (there became $6$ babies in total) and then the nurse randomly picked up a boy from among them. Hence:
$$P(B_1)=\frac12; P(G_1)=\frac12;\\
P(B_2|B_1)=\frac36; P(G_2|B_1)=\frac36; P(B_2|G_1)=\frac26; P(G_2|G_1)=\frac46.$$
Now you want to find (using Bayes' theorem):
$$P(B_1|B_2)=\frac{P(B_1\cap B_2)}{P(B_2)}=\frac{P(B_1)\cdot P(B_2|B_1)}{P(B_1)\cdot P(B_2|B_1)+P(G_1)\cdot P(B_2|G_1)}=\\
\frac{\frac12\cdot \frac12}{\frac12\cdot \frac12+\frac12\cdot \frac26}=\frac3{5}.$$
A: I would like someone to verify this, since my confidence isn't high.
This is a method that helps corroborate chances. If something doesn't quite add up, you'd generally notice, since then one or more of the equations would be wrong.
cube chart
The description explains several events with 4 conditions; the chance a boy or a girl is added, and the chance the nurse picks a boy or a girl.
What you're asking is basically, what are the odds something happens, given that we know that thing already happened. In this case it's "What are the odds a boy was added, given that we know the nurse picked a boy". You can see the events cross together with the answer being $1/4$
A: Let:
A - chosen child is a boy
N - new child is a boy
You are asked $P(N|A)$
Which is by using the bayes theorem:
$$P(N|A) = \frac{P(A|N)P(N)}{P(A)} = \frac{3/6 * 1/2}{1/2*3/6 + 1/2*2/6} = 0.6$$
We know $P(A|N)$ = 3/6 because if the new child is a boy we would have 6 childs and 3 boys, we can do the same for $P(A|!N) = 2/6$
We can calculate $P(A)$ as $P(N)P(A|N) + P(!N)P(A|!N)$ basically the probability that A happens if N happens or !N happens.
