Relative Frequencies of Payment Methods for Purchases

[Shackleford]

Homework Statement

[/B]
Relative frequencies of amount purchased and method of payment

Cash Credit Debit
$<20 .09 .03 .04
$20-$100 .05 .21 .18
>$100 .03 .23 .14(a) What proportion of purchases are paid for in cash?

(b) Given that a purchase is for more than $100, what is the probability that it is paid for by credit?

(c) Are payment by credit and amount > $100 independent events?

Homework Equations

Conditional probability, independence, etc.

The Attempt at a Solution

(a) .09 + .05 + .03 = .17

(b) P(credit | >$100) = P(credit ∩ >$100)/P(>$100) = .23/.40 = .575

(c) Independent if P(credit ∩ >$100) = P(credit)P(>$100) = .23.

P(credit)P(>$100) = 0.47*0.40 = .188. not independent.Does that look right?

 

[andrewkirk]

It looks right to me.

 

[Shackleford]

It looks right to me.

Thanks. I initially overcomplicated it and then thought about it a bit more carefully.

I have another question. Is this a true relation P(~E | F) + P(E | F) = 1?

 

[Shackleford]

Actually, I had a typo in there. It should be right now.

 

[andrewkirk]

You can reason that one out using the measures:
$$P(E| F)\equiv\frac{P(E\cap F)}{P(F)}$$
So
$$P(E| F)+P(\sim E| F)\equiv\frac{P(E\cap F)}{P(F)}+\frac{P((\sim E)\cap F)}{P(F)}
=\frac{P(E\cap F)+P((\sim E)\cap F)}{P(F)}
$$

The two sets in the numerator are disjoint, so you can use the rule for the probability/measure of the union of two disjoint sets to convert the numerator to a single P(something).