Probability of winning a ticket with a red dot Question:
You have obtained some interesting information about the local lottery. There was a malfunction at the printer that accidentally marked a bunch of tickets with a red dot. This malfunction disproportionately affected winning lottery tickets. In total $40\%$ of winning tickets were marked with a red dot, while only $20\%$ of losing tickets were marked with a red dot. You have a probability of $\frac{3}{39}$ of winning the lottery.
You have found a ticket  marked with a red dot. What is the probability that this is a winning ticket?
What i have done is the following:
A= probability of a winning ticket.
B= probability of having a red dot.
P(A|B)= P(A intersection B) / P(B)
P(B)= P(red dot| winning ticket)$\times$ P(winning ticket) + P(red dot| losing ticket)$\times$ P(losing ticket)
$0.4 \times \frac{3}{39} + 0.20 \times \frac{12}{13}= 0.215$
P(A intersection B)= P(A|B)$\times$P(B)= $0.4\times 0.215=0.086$
So P(A|B)= $\frac{0.086}{0.215}=0.4$
I'm not sure if I'm doing this right. Maybe someone can give feedback.
 A: Imagine a total of $1300$ tickets.  $\frac{3}{39}= \frac{1}{13}$ of them, $100$, are winning tickets, $1200$ are not.  $40\%,  40$, of the winning tickets have a red dot.  $20\%, 240$, of the non-winning tickets have a red dot.
That is a total of $40+ 240= 280$ tickets that have a red dot of which $40$ are winning tickets.  If you have a ticket with a red dot, the probability that it is a winning ticket is $\frac{40}{280}= \frac{1}{7}$, approximately $14\%$.
A: Out of $39$ tickets, $3$ are winning tickets and $36$ are losing tickets.
So, number of winning tickets with red dot = $3\times 0.4$
Number of losing tickets with red dot = $36\times 0.2$
You need to find probability of winning of a ticket with red dot = $\dfrac{3\times 0.4}{3\times 0.4+36\times 0.2} = \dfrac{1}{7}$
A: Apply Bayes' rule:
$$
\begin{aligned}
P(\text{winning}|\text{red})
&=\frac{P(\text{red}|\text{winning})P(\text{winning})}{P(\text{red}|\text{winning})P(\text{winning})+P(\text{red}|\text{not winning})P(\text{not winning})}\\ &=
\frac{\frac{4}{10} \cdot \frac{3}{39} }{\frac{4}{10} \cdot \frac{3}{39} +\frac{2}{10} \cdot \frac{36}{39} }\\
&=\frac{4 \cdot 3 }{4 \cdot 3 +2\cdot 36 }\\
&= \frac{1}{7}
\end{aligned}$$
