What are some ways in which companies can attract and retain employees? (site 2)
2 answers:
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Answer:
b) No, it's not independent.
c) 0.02
d) 0.59
e) 0.57
f) 0.5616
Step-by-step explanation:
To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)
Let's call A the event that she passes the first course, then P(A)=.73
Let's call B the event that she passes the second course, then P(B)=.66
Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98
b) Is the event she passes one course independent of the event that she passes the other course?
Two events are independent when P(A∩B) = P(A) * P(B)
So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:
P(A∪B) = P(A) + P(B) - P(A∩B)
We are going to solve for P (A∩B)
.98 = .73 + .66 - P(A∩B)
P(A∩B) =.73 + .66 - .98
P(A∩B) = .41
Now we will see if the formula for independent events is true
P(A∩B) = P(A) x P(B)
.41 = .73 x .66
.41 ≠.4818
Therefore, these two events are not independent.
c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)
1 - P(A∪B) = 1 - .98 = .02
d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)
1 - P(A∩B) = 1 -.41 = .59
e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.
P(A∪B) - P(A∩B) = .98 - .41 = .57
f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)
P(B|A) = P(A∩B) / P(A)
P(B|A) = .41 / .73 = .5616
