Question

I QUESTION 15 (a) Assumes that one out of ten customers entering a book shop buy books. If 10 customers visited the shop in one day, what is the probability that () One of the ten bought books, (i) Between two and five bought books, (li) No-one bought books. (b) Messages arrive at a reception desk in two different ways: via phone calls and via faxes. Assume that phone calls arrive on average at the rate of 2 per hour, and faxes on average at the rate of 4 per hour. The phone calls and faxes arrive independently of each other, and both can be modelled using the Poisson distribution (1) Find the probability that three phone calls arrive during two hours. (i) Find the probability that during one hour, there is one phone call and one fax.(5) (ii) Find the probability that there are no messages at all during a two-hour period. (2) (iv) Find the expected value and standard deviation of total number of all messages (faxes or phone calls) during one hour 122

Answer 1

1 in the given problem No.of customers visited the shop= n=10

Let us suppose that no.of visitors bought books is denoted with X and is follows binomial distribution with P= probability of success= probability of a customer bought the books= 1/10

Q= 1- P = 9/10 = probability of failure = probability of a customer not bought the books

a) probability that one of the ten customers bought the books by binomial probability function = P(X=1) = 10_C1 (1/10)¹(9/10)⁹

= 0.3874

b) probability that between 2 and 5 bought the books= P(2<=X<=5)= P(X=2)+ P(X=3)+P(X=4)+P(X=5)

where P(X=2)= 10_c2 (1/10)²(9/10)⁸= 45x0.01x0.4305=0.1937

P(X=3)=10_c3 (1/10)³(9/10)⁷= 120x0.001x0.4783=0.0574

P(X=4)= 10_c4 (1/10)⁴(9/10)⁶= 210x0.0001x0.5314=0.0111

P(X=5)= 10_{c5 (1/10)}⁵(9/10)⁵= 252x0.00001x0.5905= 0.001488

p(2<=X,<=5)= 0.1937+0.0574+0.0111+0.001488=0.26368 (if both 2 and 5 inclusive)

probability that between 2 and 5 customers bought the books= 0.2636 ( if both 2 and 5 are inclusive

or probability that between 2 and 5 customers bought the books= P(X=3)+P(X=4)= 0.0685 (if both 2 and 5 are exclusive)

c)Probability that no one bought books= P(X=0)= 10c0 (1/10)0 (9/10)10= (9/10)10= 0.3486

2.Let us suppose that arrival of no.of phone calls X follows poisson distribution with mean $\Lambda$ ₁=2

Let us suppose that arrival of no.of faxes Y follows poisson distribution with mean $\Lambda$ ₂=4

also phone calls and faxes are independent

a) probability that 3 phone calls during 2 hours= P(X=3) = e^-4 (4³)/3! =0.018315x64/6=0.1953

here in two hours average calls =4= $\Lambda$ 1

b) probability that during one hour 1 phone call and 1 fax

probability that arrival of 1 phone call= P(X=1)= e^-2 (2¹)/1! = 0.1353x2=0.2706

probability that arrival of 1 fax = P(Y=1)= e^-4 (4¹)/1! =0.018315x0.07326

we know that arrival of phone calls and faxes are independent

hence probability of getting 1 phone call and 1 fax in one hour= P(X=1)XP(Y=1) = 0.2706X0.07326=0.019824

c. probability that there are no messages during a hour= P(X=0)XP(Y=0) =0.0024787

D) mean of all messages in 1 hour = E(X+Y)= E(X)+E(Y)=2+4=6

S.D OF Total no.of messages= s.d(X+Y)=SQUARE ROOT OF(V(X+Y)0= SQUARE ROOT OF (V(X)+V(Y))= 3.4142