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Assume that the probability that an American woman has childrenduring her lifetime is 80%. Also assume that the probability thatan American woman who has children g...

Question

Assume that the probability that an American woman has childrenduring her lifetime is 80%. Also assume that the probability thatan American woman who has children gets breast cancer during herlifetime is 0.08, while the probability that any randomly selectedAmerican woman gets breast cancer is 0.11.What is the probability that an American woman both has childrenand experiences breast cancer during her lifetime? Andis having children and having breast cancer independentevents?

Assume that the probability that an American woman has children during her lifetime is 80%. Also assume that the probability that an American woman who has children gets breast cancer during her lifetime is 0.08, while the probability that any randomly selected American woman gets breast cancer is 0.11. What is the probability that an American woman both has children and experiences breast cancer during her lifetime? And is having children and having breast cancer independent events?



Answers

Assume that the probability that an American woman has children during her lifetime is 80%. Also assume that the probability that an American woman who has children gets breast cancer during her lifetime is 0.08, while the probability that any randomly selected American woman gets breast cancer is 0.11. What is the probability that an American woman both has children and experiences breast cancer during her lifetime? And is having children and having breast cancer independent events?

Hello everyone. So we'll be going to solve question number 95 from chapter two of probability. So indication has been given that we have Like randomly selected certain people from above 40 years of age and they have probably of cancer at 0.05. So let's say P. C. Is the probability of having cancer like the person has cancer. And pd let the probability of that person is diagnosed diagnosed with cancer. Notice given PC as 0.05. The probability of person having cancer is 0.05. So p person not having cancer c note will be equal to 1 -0.05. That would be 0.95. Now in the next part is given That the person has cancer and dr correctly diagnosed that particular person. That properties .78 person is diagnosed and person has cancer. This B. D. Slash E. Means the person has cancer and it is given that person has cancer and the doctor has diagnosed that person with cancer correctly. So that gravity is 0.78. It is given indication the probability that person is diagnosed with cancer. But given that person didn't have it had the cancer, the pd slash seen not. It has given us .06. This is the wrongly classification done by the doctors. The person doesn't have cancer, but doctor diagnosed that person with cancer. It's .06. Now we have to find out uh p probability of diagnosed that the doctor diagnosed a particular person of having a cancer. B. B. Will be equal to BC intersection. D. Person has the cancer and diagnosed with cancer. Plus person doesn't have cancer, but diagnosed doctor diagnosed that person with cancer. It will be so probably of diagnosis is will be the sum of these two. Know how to how to calculate pc intersection D. And B. C. Notes intersection day formula for pd slash C. This is the condition for everybody. Is that D. D. Intersection C. What we can do pc intersection day because we need that divided by probability of having see. So this is it is given that the person is cancer and we have to find the condition for everybody. The person is diagnosed. Soapy these let's see is equal to P. C. In the section D. Too worried about Pc. Using this formula with PC intersection D. Will become Bds. Let's see into PC. Probably that the person is diagnosed given that person is cancer and diagnosed with cancer and manipulation by the problem D. Of person really has a cancer. Similarly the formula for other part would be P. D. Slash C. Note multiplication by pc. No. Now we have all these values and given indication as I have written her. So we will just input the values 0.788- 0.05. 0.07 18-0.05. This will give us the remedy of person uh diagnosed doctor diagnosed that person with cancer. The next one would be 0.6 and 20.95 point 95 persons doesn't have cancer by dr incorrectly diagnosed them with cancer. So this will also come into the addition part. It would be just the calculation left for this question. 0.0 to 9 0.057 0.096 will be the probability of diagnosis, the doctor diagnosed a person with cancer. This is our final answer. Thank you. Hope you understood.

All right. So for number 18, we're now talking about the probability of females living to be 41 that's a little bit of a different probability. From the last problem, it's gonna be 410.99855 So, for part a one, another probability of two females living that till 40 41. So I'm gonna take 0.99855 and we're going to square that raise that to the second power, and that's going to be 0.9971 Approximately for be. The probability of five women living to that age would be 50.99855 to the fifth power, which is approximately point 99 277 And lastly, for part C. Probability of at least one, um of the five dying would be one minus 10.99 to 77 which is 0.0 7 to 3. And since that is less than 5% yes, that would be an unusual occurrence

Here, we can assume two events. The first ways events A You close to, um, America has traveled. Who? Canada. And the probability off a was too Your old points. 18. We have human event B is American Traveled to Mexico and the probables You be you who's there were points one night and also what used to be a dying world to help those built the problem. Oh, this is events A and they says events be, uh, sorry. Let me with Thanks year and the intersection is event a and even be And we know the, uh bearing his travel bulls comfrey a in close to they will do well, the first questions. What's ah probably t that nurtures in has trouble to Canada but not Mexico, which is a pro variety A and not B Oh, it should be these area. So we use a probable once a minus pulled a n you host Teoh their points 18 minors single 180.4. My answer. Is there a point? 14. The question bree is, uh, and marriage Hewson at random has traveled to eyes in Canada. Old Mexico. So what? We won't iss, uh, America travel to I the cannons up. Yes, travel to Canada. But it's not Mexico. Plus the probability America traveled to Mexico but not Canada. And this one we can use here is there old 0.1 Plus, similarly, these probability be and not a equals two probably want to be minus probable T B, which equals to their 0.14 plus. They will point. There were lying miners. They will point Therefore, the final readout iss their own 0.90. The questions see is ah, Marican shoes at random has not travel to idle countries, which is the every year outside the circle. So what we want is the probable Auntie um no. Traveled as a country. Don't a end not be because to for what she the total Probably one minus that probe old she a b There's equals war miners. Uh oh. This one We can use the general It is We're go. It's probable. A loss for to be minus approach A equals two world miners There were points 18 plus there were 1.9 miners every point. There were four. And we can calculate this. The finery, though, should be they will 0.7 77

Question number 11, suppose that 2% of the clinics patients are known to have cancer. A blood test has developed, that is positive and 98% of patients with cancer but is also positive in 3% of patients who do not have cancer. If a person who has chosen at random from the clinics patients has given the test and it comes out positive. What is the probability that a person actually has cancer? So, the question we're ultimate looking for is what is the probability that the person has cancer? But there's a condition to this given that they have a positive test. So, the probability of cancer given a positive test. This is the probability that someone has cancer and gets a positive test divided by the probability of a positive test. So, going back and looking at the numbers they gave us, 2% of the clinics patients are known to have cancer. Someone write that out. So the probability of cancer here, His 0.2 a blood test is developed that is positive and 98% of patients with cancer. So positive given cancer, That's how that's written. That's 0.98 and it's also positive and 3% of patients who do not have cancer. So the positive, we'll say no cancer or you can write cancer with the compliments on and that's 0.0 three. So at this point, we have everything we need to solve this problem. And I'm gonna do this by making a tree diagram. So we're gonna start off pretty simple. We're gonna say see for cancer and see compliment for not cancer. We have 02 representing cancer. That means 20.98 represents non cancer. In this scenario for patients who have cancer, they get either a positive or negative. It tells us that if they have cancer, positive test occurs 0.98% our 0.98 which is 98%. So a negative must be 980.2 For those without cancer, a positive test occurs 3% of the time. So that means a negative must occur 30.97% of the time. Now we can use this tree diagram to find the pieces that we need. So the probability of cancer and a positive test, we're looking for the numerator here, so we want to stay on the track of cancer and positive test. So that would be 0.2 tom's point 98 and then positive test. There's two different ways to get a positive test. We can get a positive test on the top of this tree, or we can get a positive test below here on the bottom. So there's two different paths past. Number one is the one we already circled. I'll circle it again. So we could have a positive test and have cancer. Or we could have a positive test but do not have cancer. But both of these results in a positive test. At this point, you can see the answer choices and answer choice E matches what we have set up in the video.


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