SOLUTION: A Cape Town courier service promises that 80%of Johannesburg-bound parcel deliveries will reach their destinations within 12 hours. What is the probability that of the seven parcel

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Question 1122069: A Cape Town courier service promises that 80%of Johannesburg-bound parcel deliveries will reach their destinations within 12 hours. What is the probability that of the seven parcels sent at random times by a particular client in Cape Town, only one is delivered late?
I worked that out and got an answer of 0,3670. I just need help with the second part which is...
What is the probability of the seven parcels sent at random times by a particular client in Cape Town, only one is delivered within 12 hours?
A) 0.0004
B) 0.2753
C) 0.2097
D) 0.0043
E) 0.3670

Thank You
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
i believe this is a binomial probability type problem.
the formula is p(x) = p^x * q^(n-x) * c(n,x)

p is the probability of success.
q is the probability of failure = 1 - p
x is the number of successes
n-x is the number of failures
c(n,x) is the number of ways you can get sets of x out of a set of n where order is not important.

p = .8
q = 1 - .8 = .2
x = 0 to 7
n = 7

the probability that only one is late is 1 minus the probability that 6 are on time.

that would be p(6) = .8^6 * .2 * c(7,6) = .3670016

the probability that only one is delivered on time would be:

p(1) = .8^1 * .2^6 * c(7,1) = 3.584 * 10^-4 = .0003584.

round to 4 decimal places and it equals .0004.

all the probabilities are listed below.

the sum of all probabilities is equal to 1 as it should be.

c(n,x) = n! / (x! * (n-x)!)