At an amusement park, the probability that a child eats popcorn and cotton candy is 0.58. The probability that a child eats popcorn is 0.69, and the probability that a child eats cotton candy is 0.87. What is the probability (rounded to the nearest hundredth) that a child eats cotton candy given that the child has already eaten popcorn?

Question

mathmate · Answer

A Venn diagram helps to explain, but not the only way to solve the problem.
|dw:1470916358775:dw|
From the diagram, it can be seen that the probability of a child eating popcorn ONLY is 0.69-0.58=0.11 and eating both is 0.58.
Given that the child has already eaten popcorn (and assuming eating anything is an independent event), then we focus on the left circle (popcorn), since outside the circle, the child does not eat popcorn.  Out of the total of 0.69, the child has a probability of 0.58 of having cotton-candy. Therefore, the probability of the child having cotton candy after eating popcorn is Pr(C|P)=(0.58/0.69), read as 
"probability of eating cotton candy given the child has eaten popcorn is 0.58/0.69."
I'll leave it to you to evaluate the expression 0.58/0.69.

Mathematically,
P(C|P) is defined as $Pr(C|P)=Pr(C\cap P)/Pr(P)=0.58/0.69$