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Homework Statement
Hi. I've been reading my probability book and I came across this problem:
In a class of 100 students, 50 have MasterCards, 60 have Visas, and 30 have neither.
Let ##V## be the event of having a Visa and ##M## be the event of having a MasterCard.
a) How many students have both cards?
b) How many students have Visa only?
c) How many students have at least one of the cards?
d) Are ##M## and ##V## disjoint?
Homework Equations
##P(A \cup B) = P(A) + P(B) - P(A \cap B)##
The Attempt at a Solution
This is my first experience with formal probability, so I will show you what I have come up with, and hopefully it is reasonable.
a) How many students have both cards?
This question reduces to asking what is ##P(V \cap M) * 100##? Re-arranging the relevant equation:
$$P(V \cup M) = P(V) + P(M) - P(V \cap M)$$
$$P(V \cap M) = P(V) + P(M) - P(V \cup M)$$
Now we know ##P(V' \cap M') = P((V \cup M)') = 0.30 \Rightarrow P(V \cup M) = 0.70##. We also know ##P(V) = 0.60## and ##P(M) = 0.50##. Therefore:
$$P(V \cap M) = 0.60 + 0.50 - 0.70 = 0.40$$
Hence 40 students have both cards.
b) This question reduces to asking what is ##P(V \cap M')##?
So we know:
$$P(V \cup M) = P(M) + P(V \cap M')$$
$$P(V \cap M') = P(V \cup M) - P(M) = 0.70 - 0.50 = 0.20$$
So 20 students should have a visa only.
c) This question reduces to asking what is ##P(V \cup M)##?
From earlier ##P(V \cup M) = 0.70##. So 70 students have at least one card.
d) No ##V## and ##M## are not disjoint because ##V \cap M \neq \emptyset##.