I was hoping someone could tell me if I did the parts to this problem correctly... thanks so much!
A more detailed examination of the company records indicates that 95% of the customers who pay one monthly bill in full will also pay the next monthly bill in full; only 10% of those who pay less than the full amount one month will pay in full the next month. (the previous problem had stated: the utility company finds that 70% of its customers pay a given monthly bill in full)
a. find the probability that a customer selected at random will pay two consecutive months in full
I just figured the problem gave me the answer, which I am guessing is .95 but to show my work I wrote:
P( full | full)
P (A | B)
P (A and B)/P(B) = .95*.7/.7 = .95
and to verify part b:
b. find the probability that a customer selected at random will pay neither of two consecutive months in full.
assuming I answered A correctly, wouldn't it just be the compliment... so
1- P(a customer pays in full both months)
1- .95 = .05
and part c:
c. find the prob. that a customer chosen at random will pay exactly one month in full.
Again, I thought that the 10% given in the problem was the answer to this question, but I have a feeling it can't be that simple. So I thought of doing it like this:
P (full | partial)
and I would use the formula: P ( A | B) = P (A and B)/P(B)
so, P (.7 | .3) = P (.7*.3)/.3 = .7
I am not sure if I am way off or not... any feedback would be really helpful! Thanks!
A more detailed examination of the company records indicates that 95% of the customers who pay one monthly bill in full will also pay the next monthly bill in full; only 10% of those who pay less than the full amount one month will pay in full the next month. (the previous problem had stated: the utility company finds that 70% of its customers pay a given monthly bill in full)
a. find the probability that a customer selected at random will pay two consecutive months in full
I just figured the problem gave me the answer, which I am guessing is .95 but to show my work I wrote:
P( full | full)
P (A | B)
P (A and B)/P(B) = .95*.7/.7 = .95
and to verify part b:
b. find the probability that a customer selected at random will pay neither of two consecutive months in full.
assuming I answered A correctly, wouldn't it just be the compliment... so
1- P(a customer pays in full both months)
1- .95 = .05
and part c:
c. find the prob. that a customer chosen at random will pay exactly one month in full.
Again, I thought that the 10% given in the problem was the answer to this question, but I have a feeling it can't be that simple. So I thought of doing it like this:
P (full | partial)
and I would use the formula: P ( A | B) = P (A and B)/P(B)
so, P (.7 | .3) = P (.7*.3)/.3 = .7
I am not sure if I am way off or not... any feedback would be really helpful! Thanks!