1. Failures of the braking system of a car occur at random, that is, the number of failures is a Poisson variable with, on average, one failure in 200000miles. find the probability that:
a) the car completes 50000miles without a brake failure
b) there are more than 2 failures in 50000miles
[Hint: you may assume that the mean number of failures is proportional to the distance travelled]
ans a) 0.7788 b) 0.0022
2.A company's switchboard handles both internal and external telephone calls. Past experience shows that from 10am to 11am, the number of internal and external calls are independent and gave Poisson distributions with means 3 and 5, respectively. find the probability that during 10 am to 11am of a particular day, there will be
a) no call at all
b) at least three internal calls
c) two internal calls and one external call
ans
a) 0.0003
b_ 0.5768
c_ 0.0075
thx!!
F.7 Probability [20 points]
2011-10-26 7:15 am
回答 (2)
2011-10-27 4:16 am
✔ 最佳答案
1a) Travelling distance = 50000 mile, we have:λ = 1/4Hence the probability that without a brake failure is:exp(-λ) = 0.7788b) P(more than 2 failures) = 1 - exp(-λ) - λexp(-λ) - λ^2 exp(-λ)/2= 0.0022
2a) P(no int. call) = exp(-3)P(no ext. call) = exp(-5)Hence P(no call) = exp(-3) x exp(-5) =0.0003b) P(at least 3 int. calls) = 1- exp(-3) - 3exp(-3) - 9exp(-3)/2 = 0.5768c) P(2 in. calls and 1 ext call) = 9exp(-3)/2 x 5exp(-5) = 0.0075
2011-10-26 5:01 pm
1a) With travelling distance = 50000 mi, we have:
λ = 1/4
Hence the prob that no brake failure is:
e-λ = 0.7788
b) Prob = 1 - e-λ(λ0/0! + λ/1! + λ2/2!) = 0.0143 (pls. verify)
2a) P(no int. call) = e-3
P(no ext. call) = e -5
Hence P(no call) = e-3 x e-5 =0.0003
b) P(at least 3 int. calls) = 1 - e-3(1 + 3 + 9/4) = 0.6888 (pls. verify)
b) P(2 in. calls and 1 ext call) = e-3 x 3 x e-5 x 5 = 0.0050 (pls. verify)
λ = 1/4
Hence the prob that no brake failure is:
e-λ = 0.7788
b) Prob = 1 - e-λ(λ0/0! + λ/1! + λ2/2!) = 0.0143 (pls. verify)
2a) P(no int. call) = e-3
P(no ext. call) = e -5
Hence P(no call) = e-3 x e-5 =0.0003
b) P(at least 3 int. calls) = 1 - e-3(1 + 3 + 9/4) = 0.6888 (pls. verify)
b) P(2 in. calls and 1 ext call) = e-3 x 3 x e-5 x 5 = 0.0050 (pls. verify)
參考: 原創答案
收錄日期: 2021-04-26 14:55:46
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