# Homework Help: Intro Probability

1. Jan 13, 2016

### Zondrina

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

3. 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 $V \cap M \neq \emptyset$.

2. Jan 13, 2016

### PeroK

That looks right to me. You should be able to check your answers with a Venn diagram.