# Conditional Probabilty

Problem:

When a company receives an order, there is a probability of .42 that
its value is over $1000. If an order is valued at over$1000, then there is a
probability of .63 that the customer will pay with a credit card.


A.) What is the probability that the next three independent orders will each be valued at over $1000? B.) What is the probability that the next order will be valued at over$1000 but will not be paid with a credit card?

For A, I think it should just be $.42*.42*.42$ = $.074$ ??

For B, if I let O = the probability that an order will be over $1000, and let C = the probability that if it is over$1000 then it is paid with a credit card ,

Then $$P(C|O) = {P(C \cap O)\over P(O)} = {.63 \over .42} = 1.5$$

that must be wrong since a probability must between zero and one. Right?

Another train of thought was :

$${P(C \cap O) \over P(O) } = .63$$ $$so$$ $${x\over.42}= .63$$ $$x=.264$$ $$B = 1-.63 = .37$$ Intuitively I thought that maybe if I get $P(C|O)$ then I could just do $1- P(C|O)$ to get the answer for B but now that doesn't seem right. Hints on how I should proceed?

It is a lot simpler than you are making it to be. If $O$ is the event that the order is over 1000, and $C$ is the event that the order is paid with a credit card (regardless of the order amount), then you are told that $$\Pr[C \mid O] = 0.63,$$ hence $$\Pr[\bar C \mid O] = 1 - \Pr[C \mid O] = 1 - 0.63 = 0.37.$$ Therefore, $$\Pr[O \cap \bar C] = \Pr[O]\Pr[\bar C \mid O] = (0.42)(0.37).$$ That's the mathematical way to do it. The reason why you got stuck is because you defined $C$ as a conditional event that requires $O$ to be observed first. Instead, try to define events as simply as possible, and then construct conditional events from these.
Another thing I can recommend you to do is to look at the problem in a frequentist manner. That is to say, suppose you have $10000$ customers. How many of these customers do you expect to have orders over \$1000? Well,$4200$, of course. Of these customers, how many do you expect to pay with a credit card?$4200 \cdot (0.63) = 2646$. So how many of these$4200$customers would not pay with a card? $$4200 - 2646 = 1554.$$ Thus the probability is$1554/10000 = 0.1554$. Your first answer is correct. The probability of the conjunction of three independent events is the product of their individual probabilities. Your second answer should be of the same form. However, here the probability of two dependent events must use conditional probability. $$\mathsf P(\mathbf O \cap \mathbf C^c) = \mathsf P(\mathbf O)\cdot (1-\mathsf P(\mathbf C\mid \mathbf O))$$ Where$\mathsf P(\mathbf C^c\mid \mathbf O) = 1 - \mathsf P(\mathbf C\mid \mathbf O)$is the conditional probability of the order being not paid by card when given that it is over$\$1000$.   You have that by the Law of Complements.