Conditioning and Independence of probability Three marksmen hit the target with probabilities $ \frac{1}{2}$,$ \frac{2}{3}$ and $ {\frac{3}{4}}$ respectively.
  They shoot simultaneously and there are two hits.
Who missed? Find the probabilities.
I have tried as follows:-
$P(A~\text{hits the target})=\frac{1}{2}= P(A)$.  Similarly, $P(B)=\frac{2}{3}$ and $P(C)= \frac{3}{4}$.   
Now, $P(A~\text{does not hit}) = P(A')= 1- \frac{1}{2}= \frac{1}{2}$.
  Similarly, $P(B')= \frac{1}{3}$ and $P(C')= \frac{1}{4}$.
Since all of them shoot but only two hit so we have three cases as follows:
$$P(A~\text{and}~B~\text{hit but}~C~\text{does not})=P(ABC')= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} = \frac{1}{12}$$
Similarly, 
$$P(AB'C)= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{8}$$
and 
$$P(A'BC)= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$$
So, 
$$P(\text{they all shoot simultaneously but only two hit})= \frac{1}{12} + \frac{1}{8} + \frac{1}{4} = \frac{11}{24}$$
Therefore, 
\begin{align*}
P(A~\text{missed the target}) & = \frac{P(A'BC)}{P(\text{they all shoot simultaneously but only two hit})}\\
& = \frac{\frac{1}{4}}{\frac{11}{24}}\\ 
& = \frac{1}{4} \cdot \frac{24}{11}\\
& = \frac{6}{11}
\end{align*}
Similarly, 
$$P(B~\text{missed})=  \frac{\frac{1}{8}}{\frac{11}{24}} = \frac{3}{11}$$
$$P(C~\text{missed})= \frac{\frac{1}{12}}{\frac{11}{24}} = \frac{2}{11}$$
Please correct me wherever I am wrong, whether explanation or calculation.
 A: Let $A,B,C$ be the events that person A,B,C hit the target respectively.
Let $X$ be the event that exactly one person missed.
The probability that exactly one misses is:
$$Pr(X)=Pr((A^c\cap B\cap C)\cup (A\cap B^c\cap C)\cup (A\cap B\cap C^c))$$
Using properties of mutually exclusive events and using properties of independent events, this expands out as:
$$=Pr(A^c)Pr(B)Pr(C)+Pr(A)Pr(B^c)Pr(C)+Pr(A)Pr(B)Pr(C^c)$$
Plugging in values:
$$\frac{1}{2}\cdot\frac{2}{3}\cdot\frac34+\frac12\cdot\frac13\cdot\frac34+\frac12\cdot\frac23\cdot\frac14 = \frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{11}{24}$$
We are tasked with calculating the values: $Pr(A^c\mid X)$, $Pr(B^c\mid X), Pr(C^c\mid X)$.
Remembering the definition of conditional probability: $Pr(E\mid F) = \dfrac{Pr(E\cap F)}{Pr(F)}$ we have:
$$Pr(A^c\mid X) = \frac{\frac12\cdot\frac23\cdot\frac34}{\frac{11}{24}}=\frac{6}{11}$$
Similarly calculated, we have $\frac{3}{11}$ and $\frac{2}{11}$ for the probabilities that $B$ and $C$ were the ones who missed respectively.  Your attempt was correct.
