Poisson Process conditional probability

Question

I have the problem

"Suppose that the number of calls arriving to a telesales call centre follows
a Poisson process with intensity λ = 6 calls per hour. Each caller makes a
purchase with probability 0.4.

Suppose now  that 5 calls arrived in the first hour. What is the probability
that 3 of them made a purchase?"

I have set $X_t =$ number of calls by time $ t $ and $Y_t =$ number of purchases made by time $ t $

So I have $Y_t \sim Po(0.4(6)t)$ and I found that arrival time is exponentially distributed and from that i have the expected arrival $\mu = \frac{1}{2.4}=0.4167$

I then found $$P(Y_1=3)= 0.20914 $$

Im not really sure how to use all this information to get the answer I want, I assume the answer is $$P(Y_1=3|X_1=5)$$ but not sure what to do

Answer 1

Since you already know that $5$ people called you no longer need to use the Poisson distribution. Try answer this question: "$5$ calls arrived in the first hour. Giver that each caller has a $0.4$ probability to make a purchase, what is the probability that $3$ of them made a purchase." If you are still unable to do it I am happy to provide a full solution.