Probability

ranish293

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Im having problem with the following question related to probability, Could any one help with the solution and kindly explain me how to solve?

The proportion of American Express credit-card holders who pay their credit card bill in full
each month is 23%; the other 77% make only a partial or no payment.


(a) In a random sample of 15 customers, what is the probability that:
i. 4 customers pay their bill in full?
ii. More than 6 customers pay their bill in full?


(b) In a random sample of 17 customers, what is the probability that:
i. 4 customers pay their bill in full?
ii. No more than 3 customers pay their bill in full?
 
Im having problem with the following question related to probability, Could any one help with the solution and kindly explain me how to solve?

The proportion of American Express credit-card holders who pay their credit card bill in full
each month is 23%; the other 77% make only a partial or no payment.


(a) In a random sample of 15 customers, what is the probability that:
i. 4 customers pay their bill in full?
ii. More than 6 customers pay their bill in full?


(b) In a random sample of 17 customers, what is the probability that:
i. 4 customers pay their bill in full?
ii. No more than 3 customers pay their bill in full?
Let's start with a simpler problem. Using the same percentages of 23% and 77%

What is the probability that a single customer chosen at random will be one who pays in full each month?

What is the probability that out of three customers chosen at random, none will pay in full each month?

What is the probability that out of three customers chosen at random, exactly one will pay in full each month?

What is the probability that out of three customers chosen at random, exactly two will pay in full each month?

What is the probability that out of three customers chosen at random, all three will pay in full each month?

What is the probability that at least two of the three will pay in full each month?

What is the probability that at most one of the three will pay in full each month?

If you can answer these problems, every similar problem will follow the same pattern.
 
Thank you JeffM for your help, it could be more helpful for me if you please solve this as i'm totally new to Probability and learning through examples.
 
Thank you JeffM for your help, it could be more helpful for me if you please solve this as i'm totally new to Probability and learning through examples.
OK Let's go through the simpler problem. Then you try the original problem and show your work. That way we can see where you are going wrong if you are.

If 23% of the customers pay in full each month, then the probability of selecting such a customer if you choose one customer at random is 23%. Does that make sense?

American Express has millions of customers so the probabilities scarcely change if we choose a small sample at random. Does that make sense?

Suppose we choose three customers at random. What is the probability that all three pay their bills in full each month?

Jump to theory. The probability of a compound event, say A and B, follows this rule

\(\displaystyle P(A\ and\ B) = P(A\ if\ B) * P(B) = P(B\ if\ A) * p(A).\) This is one of the fundamental theorems of probability.

Another fundamental theorem is this: \(\displaystyle A\ and\ B\ are\ independent \implies P(A\ if\ B) = P(A)\ and\ P(B\ if\ A) = P(A)\ and\ vice\ versa.\)

For all practical purposes, the probability of choosing American Express customers with a given characteristic in a small sample is independent.

Putting those two theorems together we get:\(\displaystyle A\ and\ B\ are\ independent \implies P(A\ and\ B) = P(A\ if\ B) * P(B) = P(A) * P(B).\)

So the probability of getting three customers who all pay their bills in full each month = 0.23 * 0.23 * 0.23 = 0.012167 = 1.2167%. With me so far?

The probability of getting three customers who all do not pay their bills in full each month = 0.77 * 0.77 * 0.77 = 0.456533 = 45.6533%. With me so far?

The probability of getting one customer who does pay the bill in full each month and two who do not is harder to figure out. The problem is that there are really three possibilities, the first person selected is the one who pays in full, or the second person selected is the one who pays in full, or the third person selected is the one who pays in full. So the probability = 3 * 0.77 * 0.77 * 0.23 = 0.409101 = 40.9101%. Do you follow that? That is the key concept.

The probability of getting two customer who pay their bills in full each month and one who does not follows similar logic. The probability is
3 * 0.77 * 0.23 * 0.23 = 0.122199 = 12.2199%.

Notice 1.2167 + 45.6533 + 40.9101 + 12.2199 = 100.0000. That is a check that we did the process correctly.

Did you understand that? If not, please tell me where I lost you.
 
Last edited:
Thanks for your reply.

So, according to what you explain, the probability would be:

(a) In a random sample of 15 customers, what is the probability that:
i. 4 customers pay their bill in full?

15 * 0.23 * 0.23 * 0.23 * 0.23 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 = 0.002368 = 0.2368%


ii. More than 6 customers pay their bill in full?

let suppose 7 customers pay their bill in full then:

15 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 = 0.0000631 = 0.00631%

correct me if i'm wrong




 
Thanks for your reply.

So, according to what you explain, the probability would be:

(a) In a random sample of 15 customers, what is the probability that:
i. 4 customers pay their bill in full?

15 * 0.23 * 0.23 * 0.23 * 0.23 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 = 0.002368 = 0.2368%


ii. More than 6 customers pay their bill in full?

let suppose 7 customers pay their bill in full then:

15 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 = 0.0000631 = 0.00631%

correct me if i'm wrong
That is incorrect. By chance are you learning about the Binomial Distribution? Because this problem is exactly that. \(\displaystyle P(k)=_{n}C_{k}(p)^k(1-p)^{n-k}\)

 
Thanks for your reply.

So, according to what you explain, the probability would be:

(a) In a random sample of 15 customers, what is the probability that:
i. 4 customers pay their bill in full?

15 * 0.23 * 0.23 * 0.23 * 0.23 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 = 0.002368 = 0.2368%


ii. More than 6 customers pay their bill in full?

let suppose 7 customers pay their bill in full then:

15 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.23 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 * 0.77 = 0.0000631 = 0.00631%

correct me if i'm wrong
Ahh I made the example too simple.

ASSUMING INDEPENDENCE of trials, the probability of success on a SINGLE trial is p, and q = 1 - p:

\(\displaystyle P(exactly\ k\ successes\ out\ of\ n\ trials) = \dfrac{n!}{k! * (n - k)!} * p^k * q^{(n - k)} = \dbinom{n}{k} * p^k * q^{(n - k)}.\)

This is a really important formula. You need to understand it and then memorize it. Do you understand it?

In the case of this problem, "success" means "pays bill in full each month." So p = 0.23 and q = 1 - 0.23 = 0.77.

You are sampling 15 so n = 15. The question asks about the probability of exactly four "successes."

So the formula gives:

\(\displaystyle P(exactly\ 4\ successes\ out\ of\ 15\ trials) = \dfrac{15!}{4! * 11!} * 0.23^4 * 0.77^{11} \approx \dfrac{15 * 14 * 13 * 12}{4 * 3 * 2} * 0.0028 * 0.0564 \approx 0.2156 = 21.56\%.\)

With me so far?

Edit The formula given by srmichael and mine are the same with slightly different notation.
 
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