Shockwave
Dec3-04, 11:39 AM
A car company has found that BMW's are leased on the average of only one day in five. If leasing a car on one day is independent of leasing on another day..
a. What is the probability distribution of the number of days between a pair of cars leased?
b. What is the probability that at most 3 days seperate a pair of leased cars?
a) For n days to occur between leasings, each of n days must have NO leasing (prob=(1-p) each) followed by 1 day of leasing (prob=p). Thus, the discrete probability distribution is given by:
P{exactly n days between leasing} = p*(1 - p)^n
To be a probability distribution, the sum over all n must equal 1, and this must be verified:
SUM{n= 0 to inf} p*(1 - p)^n = p*SUM{n= 0 to inf} (1 - p)^n
= p*{1/(1 - (1 - p))} = p*{1/(p)} = 1 (<--- VERIFIED)
where the summation formula for geometric series was used above. Since p=(1/5)=(0.2), we have for this case:
P{exactly n days between leasing} = p*(1 - p)^n
= (0.20)*(0.80)^n (=ANSWER)
b) The probability for at most 3 days between leasings is thus given by:
P{at most 3 days between leasings} = SUM{n= 0 to 3} p*(1 - p)^n
= SUM{n= 0 to 3} (0.20)*(0.80)^n = (0.2)*(1 + 0.8 + 0.64 + 0.51)
= 0.59 (=ANSWER)
a. What is the probability distribution of the number of days between a pair of cars leased?
b. What is the probability that at most 3 days seperate a pair of leased cars?
a) For n days to occur between leasings, each of n days must have NO leasing (prob=(1-p) each) followed by 1 day of leasing (prob=p). Thus, the discrete probability distribution is given by:
P{exactly n days between leasing} = p*(1 - p)^n
To be a probability distribution, the sum over all n must equal 1, and this must be verified:
SUM{n= 0 to inf} p*(1 - p)^n = p*SUM{n= 0 to inf} (1 - p)^n
= p*{1/(1 - (1 - p))} = p*{1/(p)} = 1 (<--- VERIFIED)
where the summation formula for geometric series was used above. Since p=(1/5)=(0.2), we have for this case:
P{exactly n days between leasing} = p*(1 - p)^n
= (0.20)*(0.80)^n (=ANSWER)
b) The probability for at most 3 days between leasings is thus given by:
P{at most 3 days between leasings} = SUM{n= 0 to 3} p*(1 - p)^n
= SUM{n= 0 to 3} (0.20)*(0.80)^n = (0.2)*(1 + 0.8 + 0.64 + 0.51)
= 0.59 (=ANSWER)