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An oil exploration company currently has two active projects,one in Asia and the other in Europe. Let A bethe event that the Asian project is successfuland B be the...

Question

An oil exploration company currently has two active projects,one in Asia and the other in Europe. Let A bethe event that the Asian project is successfuland B be the event that the European project issuccessful. Supposethat A and B are independentevents with P(A)= 0.3 and P(B) = 0.9.(a)If the Asian project is not successful, what is the probabilitythat the European project is also not successful?Explain your reasoning.Since the events are not independent,then A′ and B′ are mutuallyexclusive.S

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A) = 0.3 and P(B) = 0.9. (a) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning. Since the events are not independent, then A′ and B′ are mutually exclusive.Since the events are independent, then A′ and B′ are mutually exclusive. Since the events are independent, then A′ and B′ are independent.Since the events are independent, then A′ and B′ are not independent. (b) What is the probability that at least one of the two projects will be successful? (c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful? (Round your answer to three decimal places.) 12. [–/8.37 Points]DETAILS MY NOTES ASK YOUR TEACHER PRACTICE ANOTHERSuppose that the proportions of blood phenotypes in a particular population are as follows: A B AB O 0.40 0.14 0.05 0.41 Assuming that the phenotypes of two randomly selected individuals are independent of one another, what is the probability that both phenotypes are O? (Enter your answer to four decimal places.) What is the probability that the phenotypes of two randomly selected individuals match? (Enter your answer to four decimal places.)



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An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let $A$ be the event that the Asian project is successful and $B$ be the event that the European project is successful. Suppose that $A$ and $B$ are independent events with $P(A)=.4$ and $P(B)=.7 .$
(a) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.
(b) What is the probability that at least one of the two projects will be successful?
(c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

Problem. 20. The null hypothesis is that B one minus B two is equal to zero, and the alternative hypothesis is is that the one minus B two is bigger than zero, where be one proportion is X one over N ones for 30/50 she's open six and B two, which is equal to extra over in two, which is 22/50 which is Oh, point or four. The pooled proportion, which is X one plus x 2 30 plus 22 over in one plus in two she is 100. She's equal toe 1000.52 toe. Find the value off until the statistics, which is equal to the P one minus B two over a square root off 4.5 to 1 minus 4.5 52 square root off one over n. One plus one over and to which is equal to 1.6 eso. The probability is equal to the probability that that or the P value is equal, that the probability that that is bigger than 1.6, which is equal to the probability that set is smaller than negative 1.6, which is equal to open toe five, 48 and the P value is bigger than open toe five. So we say it to reject the null hypothesis is so there is no sufficient evidence to support support the claim.

It's a little event Azour first digits and it's going to be zero through nine events. Beat is going to be our second digits And also going to be zero through 9 events A and B are going to be independent since one does not depend on the other. We choose zero doesn't mean that we have to choose zero again or we can if we want and our probability of choosing two of the same numbers is simply going to be one in 10 Times, one in 10 for our first choice and their second choice. So that's gonna be a 1% chance.

This problem were given a table where categorizes 250 employees of a school into ah, their tasks and their gender. So part A is the probability that their female so there are 56 plus 26 plus 18 female. So 100 out of the 2 50 uh, are female. So 100 over 2 50 That will be 0.4 as the probability uh, next part probability that they're female and teacher will there are 56 of them. It's a 56/2 50 and when we reduce it, turn it into a decimal, that'll be 0.224 in the next one, Probably that they're female or not. In an administrative rule, there are only 14 Mehlis administrators. So if we just take the compliment to 15 minus 14 that'll be 236 people that are females or non administrators. Ah, over 250 and that'll equal to 0.944 and ah, last one in port day. Ah, probability that they are ah, not female and an administrator over the probability of being an administrator so not female and administrator would be really male and administrator, so 14 over 2 50 and probability of being an administrator. 14 plus 26. There are 40 of them, so that will be times 2 50/40 when you reciprocate the fraction because of the division. So that will be 14/40 and that will give us 0.35 in part B. We want to identify some events that are independent to F so independent to being female and mutually exclusive to being female. Ah, the mutually exclusive one is a easier one to look at in this table. Given the information, being male ah is mutually exclusive because there are no overlaps. Male, because no overlaps and independent would require a little bit more work. You want to make sure that it satisfies the condition for independent events that if you actually tested this out, the probability of f anti is actually equal to of the probability of being female times the probability of being a teacher and since the satisfies team is the event that we want, so being a teacher is independent to being a female in part C, we're being told that 90% of the teachers as well as 80% of the administrative staff and 30% of the support staff phones, cars find the probability that a staff member chosen at random owns a car. So, ah, of the, um how many teachers there are there in 84 times 56 84 plus 56. And we know 90% of them have cars. We know there are 126 teachers, so maybe I'll back up a little bit here and say, Ah, 140 times 0.9. 90% of the 140 teachers there are 40 administrators and 80% of them own cars and are 70 support staff and 30% of them own cars. Ah, out of 250 employees Just got to figure out what the numbers are in the numerator. So 1 40 times 0.9 is 1 26 40 times 400.8 This 32 and 70 times point three is 21 over 2 50 So this works out 21 79/2 50. And that is ah, 0.716 as thes probability and part two, we want to find the probability that it's a teacher, Um, given that they already own a car. So ah, on the top, that will be probability of teacher and car over the probability of owning a car, probability of being a teacher and owning a car that would be 1 40 time 0.9 and probability of owning a car is we just calculated 0.716 So that will give us, uh, one. Hang on 1 40 times zero point 9/2. 50. Sorry, because we want the probability 1 40 times 400.9 divided by 2 50 and then divided by 0.716 That will give us 0.7 04

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.


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