Well, there's three types of chocolate at play here: white, dark, and milk.
For (7): the probability that a child sampled at random likes white chocolate is one minus the probability that a child sampled at random dislikes white chocolate. We're given three probabilities: 10% of the children like only white chocolate, 30% of the children like only dark chocolate (i.e. they dislike white and milk chocolate) and the rest (60%) like all chocolates. The percentage of children who dislike white chocolate, therefore, is 1 - 0.3 = 0.7.
For (11): let's write the probabilities that we pick white, dark, and milk chocolate as W, D, and M, respectively. Let's also write the probabilities that the child likes white, dark, and milk chocolate as CW, CD, and CM.
This is a little problematic, because some of the events are not disjoint: there are, after all, kids who like all chocolates. As a workaround, let's calculate the probability that we draw a chocolate and that the kid dislikes it. Let's denote this event by K. Two possibilities here:
Luckily, most of the kids aren't picky. How can we write this probability? Well:
P(K) = P((W U M) ∩ CD) + P((D U M) ∩ CW)
Our sampling of chocolate is independent from whether the kid likes it or not, so we can write:
P(A ∩ B) = P(A)P(B)
This follows from the fact that, if A and B are independent events, then P(B|A) = P(B), and P(A|B) = P(A). That means that we can re-write P(K) as:
P(K) = P(W U M)P(CD) + P(D U M)P(CW)
We also know that W, M, and D are disjoint events: we only sample one chocolate at a time, and it's either dark, white, or milk chocolate. That means that we can write:
P(W U M) = 1 - P(D)
P(D U M) = 1 - P(W)
So we can write the previous equation as:
P(K) = (1 - P(D))P(CD) + (1 - P(W))P(CW)
Okay, so that's the probability that a trick-or-treater will dislike the chocolate we've given them. That means that the probability that they'll like it is:
1 - P(K)
