SOLUTION: Based on past experience, it is known that 10% of newly launched businesses experience cash flow problems. An MBA student selects a random sample of 5 newly launched businesses. As

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Question 848062: Based on past experience, it is known that 10% of newly launched businesses experience cash flow problems. An MBA student selects a random sample of 5 newly launched businesses. Assuming that the distribution of newly launched businesses that experience cash flow problems is independent,
a) What is the probability that at most 3 of the 5 businesses experience cash flow problems?
b) What is the probability that between 3 and 5 businesses experience cash flow problems?

Answer by swincher4391(1107) (Show Source):
You can put this solution on YOUR website!
Let X be a binomial(p=.10, n = 5)
P[X<=3]:
Two options. I suggest using the one that involves less calculations.
P[X<=3]: P[X=0] + P[X=1] + P[X=2] + P[X=3]
or
Using the complement of P[X<=3]: 1-P[X>=4] = 1-(P[X=4]+P[X=5])
I will use the 2nd.
P[X=4] = (5 choose 4) * (.10)^4 * (.90)^1 = 0.00045
P[X=5] = (5 choose 5) * (.10)^5 * (.90)^0 = 0.00001
1 - (0.00045 + 0.00001) =
Question 2:
P[3<=X<=5] = P[X=3] + P[X=4] + P[X=5]
or 1-(P[X=0] + P[X=1] + P[X=2])
The first offers less calculations, and we've already calculated 2/3 of it in the first one. All that is required is to calculate
P[X=3] = (5 choose 3) * (.10)^3 *(.90)^2 = .0081
P[X=3 or 4 or 5] = .0081 + .00045 + .00001 =
Just to make sure, I'll calculate using the complement as well.
P[X=0] = (5 choose 0)*(.10)^0*(.90)^5 = .5905
P[X=1] = 5 * .10^1 * .9^4 = .3281
P[X=2] = 10 * .10^2 * .9^3 = .0729
1 - .5905 - .3281 - .0729 = <-close enough due to some rounding