Probability questions (independent events) Three people are going to a dinner. the probability that Albertine, Karoline and Patronelle is going is 0.8,0.6,0.9 respectively.
a) what is the probability all 3 are going? is this just P(A)*P(K)*P(P)=0.8*0.6*0.9?
b) what is the probability that no1 is going? is this just P[(A/\K/\P)]'=P(A')*P(K')*P(P')?
c) what is the probability atleast one of albertine or karoline is going? is this: 
P(A/\K'/\P')+P(A'/\K/\P') + P(A/\K/\P')
d) Find the conditional probability that patronelle is going given only 1 of the three is going.
e) Given karoline is going, what is the conditional probability that albertine is also going? 
/(im pretty much clueless on d,e, not sure about c either, but I think I got a, b right, and I also think I might have to use bayes's theorem on e? or am I way off? any tips/solutions?
 A: I'll assume the three events are independent, as the title suggests. It makes sense too because without independence you can't calculate the required probabilities with the information given.
(a) Right.
(b) Your right hand side is right but not the left. It is
$$P(A^{'} \cap K^{'} \cap P^{'}) = P(A^{'})P(K^{'})P(P^{'}) = 0.2 \times 0.4 \times 0.1 = 0.008.$$
(c) You restrict the outcomes to those where Patronelle doesn't go, which the question doesn't require. There are a few ways to do this. One is:
$$P(A \cup (A^{'} \cap K)) = P(A) + P(A^{'})P(K) = 0.8 + 0.2 \times 0.6 = 0.92.$$
(d) Let $1$ be the event "only $1$ of the $3$ is going". Then,
\begin{eqnarray*}
P(P \mid 1) &=& \dfrac{P(P \cap 1)}{P(1)} \\ \\
&=& \dfrac{P(A^{'} \cap K^{'} \cap P)}{P(A \cap K^{'} \cap P^{'}) + P(A^{'} \cap K \cap P^{'}) + P(A^{'} \cap K^{'} \cap P)} \\ \\
&=& \dfrac{0.2 \times 0.4 \times 0.9}{(0.8 \times 0.4 \times 0.1) + (0.2 \times 0.6 \times 0.1) + (0.2 \times 0.4 \times 0.9)} \\ \\
&=& \dfrac{0.072}{0.032 + 0.012 + 0.072} \\ \\
&=& \dfrac{18}{29}.
\end{eqnarray*}
(e) $$P(A \mid K) = P(A) = 0.8 \qquad\text{since $A,K$ are independent}.$$
