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Intro Probability

  1. Jan 13, 2016 #1

    Zondrina

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    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. jcsd
  3. Jan 13, 2016 #2

    PeroK

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