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The problem I am currently working on is:

Consider the following information: where

  • $A$={Visa Card}
  • $B$={MasterCard}
$P(A)=0.5$, $P(B)=0.4$, and $P(A \cap B) = 0.25$

The part I am having difficulty with is part (e):

Given that an individual is selected at random and that he or she has at least one card, what is the probability that he or she has a Visa Card?

For some reason it is just eluding me. Could someone help me?

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Do you understand how to calculate conditional probability in general? –  Ben Millwood Feb 25 '13 at 0:42

2 Answers 2

up vote 2 down vote accepted

The probability of having at least one card is: $$P(A)+P(B)-P(A\cap B) = 0.65 $$

Denote C={at least one card).

The probability you need (definition of conditional probability):

$$ P(A\;|\;C) = \frac {P(A\cap C)}{P(C)}$$

If you have a Visa card, you have at least one card, so $P(A\cap C)=P(A)$

$$ P(A\;|\;C) = \frac {P(A)}{P(C)} = \frac{0.5}{0.65}$$

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Suppose there are $100$ people. How many have a Visa card? how many a Mastercard? how many both? of those who have at least one, how many have a Visa card?

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