TITLE: Probability involving unions and intersections
QUESTION [0 upvotes]: Out of a group of people, 60% like Vegetables, 70% like Chocolate, and 40% like vegetables and chocolate. Find the probability that a randomly selected person in the group neither likes Vegetables nor Chocolate.
$P(V)=0.60,$ $P(C)=0.70,$ and $P(V \cap C)=0.40$
Is this equivalent to saying $P(not(likes \; Vegetables \; OR \;Chocolate))?$
Hence $P(V^c \cap C^c)?$
Then $P(V^c \cap C^c)= P(V^c)+P(C^c)-P(V^c \cup C^c)$
Where the superscript c is the complement of each event?
Is my logic flawed?
Thanks in advance for any advice or hints.

REPLY [1 votes]: Your logic seem OK, but not the most direct route to the answer.
$P(V\cup C) = .6 + .7 - .4 = .9$
Then explain why to subtract from 1 for $P[(V \cup C)^c].$
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