Question

a store has been grossing over $850 per day about 60% of business days it is open, what are the probabilities the store will gross over $850
(a) at least 3 out of 5 days?
(b) at least 6 out of 10 days?
(c)fewer than 5 out of 10 days?
(d) fewer than 6 out of the next 20 days?

Answer 1

Since this is a binomial distribution, you have p = 0.6

Remember a binomial is (n choose k) p^k (1-p)^n-k

a) At least 3 out of 5 business days
Which means you want to find a probability when there are 3, 4, and 5 business days greater than $850 and add them up. So you have:

(5 choose 3) 0.6^3 (0.4)^2 + (5 choose 4) .60^4 (.4)^1 + (5 choose 5)0.6^5 (0.4)^0

= 0.68256

The probabilities the store will gross over $850 at least 3 out of 5 days​ ​is 0.68256

b) You want to find p(6) + p(7) + p(8) +p(9) + p(10)

which is (10 choose 6) .6^6(.4)^4 + (10 choose 7) .6^7(.4)^3 + (10choose 8) .6^8(.4)^2 + (10 choose 9) .6^9(.4)^1 + (10 choose 10).6^10 (.4)^0

= 0.63310

The probabilities the store will gross over $850 at least 6 out of 10 days​is 0.63310

c) fewer than 5 out of 10

At least 6 means we find probability when there are 6, 7, 8, 9, or10 business days that will gross over $850, fewer than 5 means 4,3, 2, 1 , 0 business days grossing over 850)

Since we already found at least 6 business days, we just need to find p(5) and then add it to P(at least 6) and subtract it from1.

(10 choose 5)(.6^5)(.4^5) = .2006

p(at least 5) = p(at least 6) + p(5) = .63310 + .2006 = .83376

1-.83376 = 0.1662

The probabilities the store will gross over $850fewer than 5 out of 10 days is 0.1662

d) This is the same as finding p(0) + p(1) + p(2) + p(3) + p(4) +p(5) where n = 20

(20 choose 0)(.6^0)(.4^20) + (20 choose 1)(.6^1)(.4^19) + .... +(20 choose 5)(.6^5)(.4^15)

= 0.0016

The probabilities the store will gross over $850 fewer than 6 out of the next 20 days is 0.0016