Application of Bayes Theorem Frank steps out from his afternoon teatime to take a phone call and leaves his tea on the table. While he is taking the call, his father walks by his tea with $P(I) =  0.15$  , his mother with $P(J) = 0.3$  , his brother with $P(K) = 0.5$  and his sister with $P(L) = 0.05$.  
When Frank comes back to his tea, he finds it filled with salt. Frank believes that whenever his father walks by his tea, he has a probability of $0.05$  to put salt in it, when his mother walks by, a probability $0.03$  , when his brother walks by, a probability $0.01$  and when his sister walks by, a probability $0.05$. 
I'm trying to find the conditional probabilities $P(I|S), \,  P(J|S), \, P(K|S), \,P(L|S)$.
So, from my interpretation, it appears that walking by is a mutually exclusive event; only one of them could have walked by. I'm also given all of the reversed conditional probabilities for the four events, so Bayes' Theorem seems like the obvious tool to select.
However, my issue is in finding $P(S|I'), \,  P(S|J'), \, P(S|K'), \,P(S|L')$.
It seems valid that, by extending the apparent mutually exclusive nature of the events, to believe that $P(S|I')$  , for example, is just $P(S|J)+P(S|K)+P(S|L)$  , leading me to believe that it was Frank's mother who placed the salt in his tea.
Is this reasoning correct?
 A: Bayes' Theorem is the right tool, but there is no need to consider the individual complementary probabilities.
Draw a tree diagram, I find it very useful.
From the apex, you have four branches, each representing the walk-by events $I,J,K,L$ with respective probabilities $p(I), p(J), p(K), p(L)$.
From each of those nodes, you can branch into $S$ (salt added) and $S'$ (no salt added). The "salt added" branches would have conditional probabilities of $p(S|I), p(S|J), p(S|K), p(S|L)$ respectively. You're only concerned with these branches since you already know the salt has been added.
So if you now want to find the individual probabilities for each family member having done it, you apply BT four times.
For the father:
$$P(I|S) = \frac{p(S|I)\cdot p(I)}{p(S|I)\cdot p(I) + p(S|J)\cdot p(J) + p(S|K)\cdot p(K) + p(S|L)\cdot p(L)}$$
You can construct analogous formulae for the other family members. Note that the denominator remains the same, so calculate that first.
I get:
$p(I|S) = 31.25\% \ (=\frac{5}{16}), \\ p(J|S) = 37.5\%\ (=\frac{3}{8}), \\ p(K|S) = 20.8\overline{3}\%\ (=\frac{5}{24}), \\ p(L|S) = 10.41\overline{6}\%\ (=\frac{5}{48})$
