Average occupancy rate at a Motel is about 40 percent.?

Question

It has 15 rooms. Assuming binomial distribution, find the probability that on any given day

a) exactly 6 rooms will be occupied
b) at least 3 rooms will be occupied
c) no more than 9 rooms will be occupied
d) the motel will be full

Answer 1

np = 0.4*15
n = 15, p = 0.4

P(X = x) = 15Cx * 0.4^x * 0.6^(15-x)

Answer 2

These are all similar questions to solve...

a) 15choose6 * (.4)^6 * (.6)^9 = 20.66%

b) at least 3 rooms = 1 - 0 rooms - 1 room - 2 rooms = 1 - (.6)^15 - 15choose1 * (.4) * (.6)^14 = 99.48%

c) no more than 9 rooms = 1 - 15 rooms - 14 rooms - 13 rooms - 12 rooms - 11 rooms - 10 rooms = 1 - .4^15 - 15*.4^14*.6 - 105*.4^13*.6^2 - 455*.4^12*.6^3 - 1365*.4^11*.6^4 - 3003*.4^10*.6^5 = 96.62%

d) (.4)^15 = 0.000107%

The trick is to realize that the probability of "x" number of rooms being occupied is
15chooseX * 0.4^X * 0.6^(15-X)

15chooseX is defined as 15! / X!(15-X)!

Good luck!

Answer 3

the probability that a room is occupied is .4 and not occupied is .6

6 rooms accupied, then 9 rooms are not.
(.4)^6 * (.6)^9

But any 6 out of 15 can be occupied. 15C6

P = 15C6 * (.4)^6 * (.6)^9
P = .2066 or 20.66%

b) at least 3 rooms are occupied. Find the probability of 2 occupied room plus 1 occupied room plus no occupied room and subtract from 1

P (at least 3) = 1 - (2 occupied room + 1 occupied room + 0 occupied room)

P = 1 - (15C2 (.4)^2 (.6)^13 + 15C1 (.4)^1 (.6)^14 + 15C0 (.4)^0 (.6)^(15)

P = 1 - (.02711)
P = .97289 or 97.289%

c) simmilar to part B. Just add the probability from 0 occupied room to 9 occupied rooms

d) (.4)^15 = .000107%

Answer 4

What ever you do, you will be working with the function (p(room occupied)=0.4)
(.4 + .6)^15, which has 16 terms
Use the first term (0.4)^15 for (d)
Use the term (0.4)^6(0.6)^9 [6!*9!/15!] for part (a)
You will have to sum terms for the other problems.

Answer 5

I might be wrong, but I think the probability of x rooms being occupied is
P(x) = (0.40)^x * (0.60)^(15-x)

a) (0.40)^6 * (0.60)^9

b) 1 - [P(0) + P(1) + P(2)]

c) 1 - [P(10) + P(11) + P(12) + P(13) + P(14) + P(15)]

d) (0.40)^15