Rock concert tickets are sold at a ticket counter. Females and males arrive at t

Question

Rock concert tickets are sold at a ticket counter. Females and males arrive at times of independent Poisson processes with rates 30 and 20 customers per hour.
(a) What is the probability the first three customers are female?

(b) If exactly two customers arrived in the first five minutes, what is the probability both arrived in the first three minutes.

(c) Suppose that customers regardless of sex buy one ticket with probability 1/2, two tickets with probability 2/5, and three tickets with probability 1/10. Let Ni be the number of customers that buy i tickets in the first hour. Find the joint distribution of (N1,N2,N3)

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Explanation / Answer

We have given that Rock concert tickets are sold at a ticket counter.

Females and males arrive at times of independent Poisson processes with rates 30 and 20 customers per hour.

Let X be the random variable is female arrive at a ticket counter.

and let Y be the random variable is male arrive at a ticket counter.

X ~ Poison(30)

and Y ~ Poison (20)

The pmf of X is,

P(X=x) = (e- * x) / x!   

and the pmf of Y is,

P(Y=y) = (e- * y) / y!

(a) What is the probability the first three customers are female?

Parameter of X is = 30 customers per hour = 30 / 60 = 0.5

P(X=3) = (e-0.5 0.53 ) / 3! = 0.0126

(b) If exactly 2 customers arrived in the first five minutes, what is the probability both arrived in the first three minutes.

Let X1 and X2 be the number of customers arriving in the first three minutes.

That means we have to find this P(X1+X2 = 2).

In that probability there are two possible cases either male or female or both arrived in first three minutes.

P(X1+X2=2) for male or P(X1+X2=2) for female or P(X1+X2=2) (for male and female)

P(X1+X2=2) for male = P(X1=1,X2=1) + P(X1=2,X2=0) + P(X1=0,X2=2)

X1 and X2 are independent.

For male the parameter is 20/60 = 0.33

P(X1+X2=2) for male =P(X1=1)*P(X2=1) + P(X1=2)*P(X2=0) + P(X1=0)*P(X2=2)

= 0.2372*0.2372 + 0.0391*0.7189 + 0.0391*0.7189  

= 0.1125

Now similarly calculate for female:

P(X1+X = 2) = P(X1=1)*P(X2=1) + P(X1=2)*P(X2=0) + P(X1=0)*P(X2=2)

= 0.3033*0.3033 + 0.0758*0.6065 +  0.0758*0.6065

= 0.09197 + 0.045985 + 0.045985

= 0.18394

hOPE THIS WILL BE HELPFUL.

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