Thirty percent of all customers who enter a store will make a purchase. Suppose that six customers enter the store and that these customers make independent purchase decisions.

a. let x= the number of the six customers who will make a purchase. Write the binomial formula for this situation.

b. use the binomial formula to calculate
1. the probability that exactly five customers make a purchase.
2. the probability that at least three customers make a purchase.
3. the probability that two or fewer customers make a purchase.
4. the probability that at least one customer makes a purchase.

Were you able to write the binomial formula?

Quote:

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Were you able to write the binomial formula?

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i don't really know if my answers are right

a. p= 0.3 of all customers make a purchase : 0.7 of all customers don't make a purchase

b1. (5)x(0.3)^5 (0.7) = 0.1 so P(5) = 0.01

b2. p(x≥3) = p(3)+p(4)+p(5)+p(6) <--- this is not complete i don't know how to complete it, i should use megastat i guess..

b3. p(x≥2) = p(2)+p(1)+p(0) <--- this one too

before. p(x≥1) = 1-p(0)

I don't really know if my answers are right

a. p= 0.3 of all customers make a purchase : 0.7 of all customers don't make a purchase

b1. (5)x(0.3)^5 (0.7) = 0.1 so P(5) = 0.01

b2. p(x is greater than or equal to 3) = p(3)+p(4)+p(5)+p(6) <--- this is not complete I don't know how to complete it, I should use megastat I guess..

b3. p(x is greater than or equal to 2) = p(2)+p(1)+p(0) <--- this one too

before. p(x is greater than or equal to 1) = 1-p(0)

Well, it seems you missed part a...

This should be:

X ~ B(6, 0.3)

b1. Good! I don't know what is the accuracy asked for, but I usually go with 3 significant figures and thus, 0.008505

b2. Good, but it would be easier if you did: 1 - (P(0) + P(1) + P(2)) I don't know what is megastat, but you can do it using the same method as in b1, by working out each of the probabilities.

b3. Well, you would already have those of P(0) and P(1) here, so you can work this out using the same method as in b2. What you posted above, is wrong though. Greater than or equal to 2, P(0) can't be, nor can P(1) nor P(2). You can remove them from probability 1, as I said in b2.

before. Yes! :)

If r options out of total n available options are supposed to be favorable, then

p = nCr x kr x (1-k)(n-r)

kr = k to the power r

(1-k)(n-r) = (1-k) to the power (n-r)


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