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Randomly selecting student at certain university. Let A denote the event the Consider Visa credit card and B be the analogous event for MasterCard sel ected indivi...

Question

Randomly selecting student at certain university. Let A denote the event the Consider Visa credit card and B be the analogous event for MasterCard sel ected individual has Suppose that P(A) = 0.75, PC)= and P(AnB) 0.65. Fill in the probabilities in the following Venn Diagram_ There are disjoint; enclosed regions representing AnB, AQBs AB; and AQB; The regions make Up the entire sample space, thus the regions have probabilities that sum to [.If the randomly selected student admits to having Mast

randomly selecting student at certain university. Let A denote the event the Consider Visa credit card and B be the analogous event for MasterCard sel ected individual has Suppose that P(A) = 0.75, PC)= and P(AnB) 0.65. Fill in the probabilities in the following Venn Diagram_ There are disjoint; enclosed regions representing AnB, AQBs AB; and AQB; The regions make Up the entire sample space, thus the regions have probabilities that sum to [. If the randomly selected student admits to having MasterCard, what is the probability that student also has Visa card? P(AIB) If the randomly selected student denics having MasterCard_ what !s the probability that student also denics having Visa card? P(4" |B*) If the student admits to having major credit card (MC or Visa), what is the probability the student has MasterCard? P(B|AU B) Note: B E(A J B)



Answers

Consider randomly selecting a student at a certain university, and let $A$ denote the event that the
selected individual has a Visa credit card and $B$ be the analogous event for a MasterCard. Suppose that $P(A)=.5, P(B)=.4,$ and $P(A \cap B)=.25 .$
(a) Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event $A \cup B )$ .
(b) What is the probability that the selected individual has neither type of card?
(c) Describe, in terms of $A$ and $B$ , the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

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

So for number one, we're looking at male and female students with long hair or with short hair. And we're just riding out the symbols for each of these problems and a through J you're gonna let f represent a female on film. A male s is gonna represent short hair, and l is gonna represent you want to go through these, eh? Probability that a student does not have long hair is just written probability parentheses. And because we're focusing on not having long hair, we're gonna right along here with an apostrophe. This is gonna indicate the compliment of long hair, which is, in essence, the same thing as the probability of short hair. Be the probability that a student is male or has short hair probability print the seas male. We're just gonna represent with the Capitol or s, which stands for short hair. You see the probability that a student is a female and has long hair? We're gonna do probability female. We can write the word and ill for long Here, Indy. The probability that a student is male given that student here, this looks a little bit differently in this case on D, the probability that a student his mail given we're gonna represent that what they line down the middle, that they have long hair. So what we're doing here is actually changing our interest from the entire population to Justus with here in e the probability that a student has long here given that the student is a mill, you're gonna flip those letters around long here. Given that we have a male, Yeah, all the female students probability that a student has short here, so it doesn't have the word given in this one. However, there indicating that that's who were interested in this female students. So of all female students, the probability that a student have short hair. So we're looking at the probability that a short hair out of all the females g of all the students with long hair. So that's what we're given the probability that a student is female. So we're looking at female given long here h the probability that a student this female or has long hair just gonna separate those with or but probability that I randomly slept. The student is a male student with short hair, So a male student with short hair indicates that he's a male and he has short hair. The probability that a student is female, it's just purely probability of.

Should resolve this question. Critical numbers typically have 16 days. It's and not all of them are random. But not all of them are. And what is the probability of randomly generating 16 digits and getting your MasterCard number? Well, there are 10 numbers. Alright, there are 16 places. There are 16 places, and on every place there are 10 numbers. So which means there are 10 raised to 16 possible combinations. It means the probability off me getting my master card number just by guessing is one by 10. Raise to six. Yes, receives often show the last four digits of the credit card number. If only those last four digits are known. What is the probability of randomly generating the other digits off your MasterCard number? Well, now I know the last four digits, which means I still have to guess that 12 remaining digits. So my probability in this case is going to be won by 10 Raised to 12. The discovered cards begin with digits 611 If you know the first four digits and you also the excuse me and you also know the last four digits. What a discover card. What is the probability off randomly generating other digits and getting all of them correct. This means I know eight visits in all I know the first four digits. And I also know the last four digits. Now there are eight. They just in the middle that I wanna guess. So what is the probability of getting them right? One divided by 10 raised to it. This is my answer for the last part. Well, is this something to worry about? Well, I wouldn't think so, because, you know, still, this is a very small probability of getting it right in the first time. But yes, I mean, it is only obvious the more digits you reveal, the higher the probability gets.

Okay. So for this problem were given several different types of probabilities and individual outcomes and were asked to find out whether or not these are legitimate sample spaces and probabilities. So the main rule that we need to remember for this particular problem is that whenever I add up all of these probabilities sum them up, it has to come out toe one. Because if I add up all the probabilities, while all the probabilities should equal 100% probability of any of those probabilities happening is 100% so that if those numbers don't add up toe one and we know it's not feasible. So let's start with this one. We have a probability of 10 plus probably too 1/6. Plus the probability of +31 of the three, probably four, was probably five, plus the probability of since which, where we add these up comes out toe one. So part A is indeed feasible. Now let's move on the part beat. Whenever I add these guys up, it's the same sort of thing. We're just adding these up get probability of, ah, full time female worker, plus the probability of a full time male worker, plus the problem of a part time female worker, plus the probability of a part time male worker. Now, whenever I add these numbers up, I get 1.41 which is not equal to one. So this one is not feasible. It's not legitimate and moving on to the last one. We're giving a shuffle deck, and we just need to add these up and hope that they add up to one. So I have probability of drying clubs, plus the probability trying a diamond plus the probability drying heart, plus the probability of Dr Spayed. And whenever I add all of these up, this adds up to one. So this is indeed legitimate and satisfies our rules of probability, and that's all all there is to it.


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