Probability of Credit Cards P(A|B')

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1. Oct 6, 2012

jegues

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

2. Relevant equations

3. The attempt at a solution

$$P(A | M^{'}) = \frac{P(A \cap M^{'})}{P(M^{'})}$$

I know that,

$$P(M^{'}) = 1-0.48 = 0.52$$

but I can't figure out how to obtain,

$$P(A \cap M^{'})$$

Any ideas?

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2. Oct 6, 2012

Simon Bridge

Isn't it given to you? Probability has a mastercard and an amex?

3. Oct 6, 2012

Dick

No, it's probability of NOT having a mastercard and having an amex. I'd suggest to jegues to draw a Venn diagram and start labelling regions.

4. Oct 7, 2012

Simon Bridge

Oh I see what you mean - did you construct the ven diagram?

$A$ = has an amex
$M$ = has a mastercard
$A\cap M$ has a mastercard and an amex.

But assign an arbitrary number of people ... say 100.

48 have a mastercard (52 don't)
20 have an american express
11 have a mastercard and an amex

Thus, 37 people with a mastercard do not have an amex.
So the probability that someone with a mastercard does not have an amex is 37/48.

That's probably the quickest way to understand the problem ... you can rework it all in terms of conditional probabilities later. (To complete this diagram: how many people have neither an amex nor a master-card?)

5. Oct 7, 2012

jegues

See figure attached for my attempt at drawing a Venn diagram.

I'm having some trouble finding the probability of not having a mastercard or a visa or a amex.(i.e. being outside of all 3 circles)

My attempt at this would be,

$$0.7 + (0.48-0.29) + (0.2 - (0.11 + 0.14 - 0.06)) = 0.9$$

but that answer doesn't make any sense.

Can someone explain how to do this? Do I need to do this to get closer to finding my answer?

EDIT: After thinking about it more,

$$P(A \cap M^{'}) = 0.2-0.11 = 0.09 \quad \quad P(M^{'}) = 0.52$$

$$P(A|M^{'}) = \frac{0.09}{0.52} = 0.17307$$

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6. Oct 7, 2012

Simon Bridge

You only need two circles like I said: see post #4.
I think that's as close as I can get to telling you how to do it without doing the problem for you.

If there are 100 people all together:
1. How many people, in total, have an amex? (given)
2. How many of the people who have an amex also have a mastercard? (given)
3. How many people who have an amex don't have a mastercard? (from 1 and 2)
4. What is the total number of people without a mastercard?
5. what is the probability that someone without a mastercard has an amex? (from 3 and 4)

 our posts seem to have crossed ... what's important is your method.
(I got 9/52 ≈ 0.17308 different rounding)

Last edited: Oct 7, 2012