Like

Report

Return to the credit card scenario of Exercise $14,$ where $A=\{$ MasterCard $\}$ $P(A)=.5, P(B)=.4,$ and $P(A \cap B)=.25 .$ Calculate and interpret each of the following probabilities (a Venn diagram might help).

(a) $P(B | A)$

(b) $P\left(B^{\prime} | A\right)$

(c) $P(A | B)$

(d) $P\left(A^{\prime} B\right)$

(e) Given that the selected individual has at least one card, what is the probability that he or she has a Visa card?

Check back soon!

Probability Topics

You must be signed in to discuss.

okay for this problem? Part A. We want the probability of being given a. It's going to be equal to the probability of a Intersect. Be divided by the probability of A, which is going to be equal to 0.25 divided by 0.5 just going to be itself 0.5 Part B. We want the probability of not to be given a well. We know that the probability of be given a and the probability of not be given a need to add upto one. You'll have either the MasterCard or not have the MasterCard 100% chance of that being true. So we can actually calculate this as one minus the probability of be given okay, which is going to be equal then to 0.5. Since we know that there's a 50% chance of you having the MasterCard given that you have a visa. Next, we want the probability of a given B, which is going to be the probability of a Intersect be divided by the probability of B, which in this case going to be 0.25 divided by 0.4, which comes out to moment here 0.5 divided by 0.4 Oh, sorry, 0.25 provided by 0.4. She comes out, too, then 0.6 to 5. And similarly, we want the probability of not a given be. It's going to be one minus the probability of be given a sorry one, minus the probability of a given be rather equal to one minus 0.65 which is equal to 0.375 Lastly, we want to find the probability of a given that we have at least one card. So the probability that we have card A, given that we have at least one card that is going to be probability of the intersection of a and one card or one plus card, I'll say, divided by the probability of one plus card. So we need to figure out what those two probabilities are. Well, the probability of a Intersect. Actually, it'll be easier to figure out the probability of one plus card first. So the probability of having at least one card is going to be the probability that we have one card or that we have a visa and not a MasterCard, that we have a MasterCard and not a Visa or that we have a Visa and MasterCard. So that's going to be the probability of be prime, given a plus. The probability of a prime given B plus the probability of a Intersect be, Um oh, actually, I need to make a slight adjustment here because we want these to be rather the intersections, not the givens be prime and a or a prime and be or A and B. We don't have those intersections given as probabilities, but we can calculate them pretty easily using Be prime, given a times the probability of a plus the probability of a prime given B times the probability of B plus. We do have the probability of a Intersect B, which is 0.25 So putting everything together be prime given A with zero point. So we'll have 0.5 times 0.5 plus. That's going to be 0.375 times ability of be used four plus 0.25 It's going to be so that totals up to 0.65 Now, the probability of the intersection of having a having a visa. And we have at least one card. Well, that No, actually, that's not an f. All right? The the of A and one plus card. What are the different possible ways we could have that Either we have a visa and no MasterCard or we have a Visa and MasterCard. So that is going to be the probability of be prime intersect. A plus probability of a Intersect be which is going to be so be prime intersect A was what this was here. So that's going to be 0.25 plus 0.25 ups, not 0.50 point 25 So that is going to be 0.5 now, putting all of this together probability that we have a and at least one card was 0.5 on the top 0.5 divided by we had 0.65 on the bottom. So that is going to give us a 0.5. Divided by 0.65 is going to give us 0.769 Approximate