Probability: Conditional Vs. Joint Issue

#1
Hi all,

Brand new to this site. I've read the posting rules, but please feel free to herd me into the correct way of doing things with this community.

I am dealing with this problem below. If you can see where my thinking is going awry, feel free to talk to me like a little kid. I want to understand more than I want an answer.

Problem
At a certain gas station, 40% of all customers fill their tanks. Of those who fill their tanks, 80% pay with a credit card. If three customers are randomly selected, what is the probability that all three fill their tanks and pay with a credit card?

My Work
Translation: P(Credit | Fill) = .8 Which means that P(Cash | Fill) = .2.
This seems like I should be finding a mixture of joint and compound probabilities. From the question, I think I want to find "P(C and F) and P(C and F) and P(C and F)."

P(C and F) should be 0.32:
P(C|F) = P(F and C) / P(F)
0.8 = P (F and C) / .4
0.32 = P(F and C)
32%

So to solve, I do this:
P(C and F) and P(C and F) and P(C and F)
= P(C and F) * P(C and F) * P(C and F)
= .32 * .32 * .32
= 3.28 %

The comment my professor left me was: Why are you multiplying them, not adding them here?

I have been all over youtube, and all over several stats sites, and while I can maybe see where I should be doing something more like, P(C|F) and P(C|F) and P(C|F), but the question isn't asking that. There are people who fill their tanks and pay with cash/checks. Can you help me figure out where my thinking went awry?

I feel like if I did what my prof told me and added them, I would get an incredibly high number (96%), but also I would be answering the question, "what is the probability of choosing three random cars where the first or the second or the third paid with credit card?"

Any help is appreciated - I hope I have shown my work without being too wordy or aggravating.

Thanks,
Dani
 

Dason

Ambassador to the humans
#2
What you did was correct. Clearly adding them wouldn't be the way to go - for example imagine if you had selected 4 customers instead of 3. If you added .32 four times you would get 1.28 which doesn't make sense for a probability.

I suppose your professor was just looking for justification as to why you did what you did. Suppose we let \(X_1, X_2, X_3\) be random variables that indicate if the customer both filled their tank and paid with a credit card (where X_1 is 1 if the first customer you selected did both and 0 otherwise, similarly for customers 2 and 3).

Then we're interested in \(P(X_1 = 1 \cap X_2 = 1 \cap X_3 = 1)\)

Since we're assuming that these events are independent (whether customer 1 filled their tank and paid with a credit card shouldn't have an impact on whether customer 2 fills their tank and pays with a credit card) we can break this probability statement up so that
\(P(X_1 = 1 \cap X_2 = 1 \cap X_3 = 1) = P(X_1 = 1)P(X_2 = 1)P(X_3=1)\) and from there it's just what you had. So the justification is really just that the events are independent and we can break the probability of the intersection of independent events into the product of the probabilities of the corresponding events.