Question 1.Customers arrive at a busy check-out counter of a

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Question 1.Customers arrive at a busy check-out counter of a large grocery store at an average rate of 6 per minute.It is assumed that number of customer arrivals follows Poisson distribution.
a.Find the probability that in any given minute there will be three or fewer arrivals.
b.Find the probability that there will be no less than 5 arrivals in two minute period

显然这里泊松分布的那个参数就是6,根据泊松分布的定义,有
P（X=k）=(6^k)*[e^(-6)]/k!
第一个问就是
P=P(X=0)+P(X=1)+P(X=2)+P(X=3)
把每一项都代入到上面的式子里,就能求出来了
第二个问
2分钟内,顾客少于5的概率：
第一分钟顾客是0,第二分钟顾客是0或1或2或3或4
第一分钟顾客是1,第二分钟顾客是0或1或2或3
.
第一分钟顾客是4,第二分钟顾客是0
把上面每一项的概率都求出来,相加,就是2分钟内顾客少于5的概率
再用1-上面那个结果就可以了.
比如求“第一分钟顾客是0,第二分钟顾客是0或1或2或3或4”的概率就是
P(X=0)*【P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)】
具体结果我就不算了,看不明白HI我
第一个问的过程：
P（X=0）=（6^0）*e^(-6)/0!=1*e^(-6)/1=e^(-6)
P(X=1)=6^1*e^(-6)/1!=6*e^(-6)
P(X=2)=6^2 * e^(-6) / 2!=18*e^(-6)
P(X=3)=6^3 * e^(-6) / 3!= 36 * e^(-6)
P=P(X=0)+P(X=1)+P(X=2)+P(X=3)=(1+6+18+36)*e^(-6)
=61*e^(-6)=0.15120388277664792

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