Question

You and a friend, along with an eccentric rich probabilist, are observing a Poisson process whose arrival rate is λ = .5 per hour. The probabilist offers to pay you $100 if there is at least one arrival between noon and 2pm, and also offers to pay your friend $100 if there is at least one arrival between 1pm and 3pm.

a. What is the probability that either you or your friend, or both, gets $100?

b. What is the probability that one of you wins $100, but not both?

Consider a Poisson process with arrival rate λ per minute. Given that there were three arrivals in the first 2 minutes, find the probability that there were k arrivals in the first minute; do this for k = 0, 1, 2, and 3.

Given that P(A) = .4, P(A ∩ B) = .1, and P((A ∪ B) c ) = .2, find P(B).

Answer 1

a)

P(that in 2 hours at least one arrival) =P(X>=1)=1-P(X=0)=1-(e^-0.5*2*(0.5*2)⁰)/0!=0.6321

hence P( both, gets $100)=P(fm noon to 2 pm one customer arrives and from 1pm to 3pm at least one customer arrives) =0.6321*0.6321=0.3996

b)

probability that one of you wins $100, but not both =P(fm noon to 2 pm one customer arrives and from 1pm to 3pm none customer arrives)+fm noon to 2 pm none customer arrives and from 1pm to 3pm at least one customer arrives)

=0.6321*(1-0.6321)+(1-0.6321)*0.6321=0.4651

c)

(i) P(K=0 arrival in first minute|3 arrival in first 2 minute)

=P(K=0 in first and 3 in second minute)/P(3 arrival in first 2 minute)

=(e^-0/0!)*(e^-3/3!)/(e^-2(2)³/3!) =1/8

ii)

P(K=1 arrival in first minute|3 arrival in first 2 minute)

=P(K=1 in first and 2 in second minute)/P(3 arrival in first 2 minute)

=(e^-1/1!)*(e^-2/2!)/(e^-2(2)³/3!) =3/8

iii)

P(K=2 arrival in first minute|3 arrival in first 2 minute)

=P(K=2 in first and 1 in second minute)/P(3 arrival in first 2 minute)

=(e^-2/2!)*(e^-1/1!)/(e^-2(2)³/3!) =3/8

iv)

P(K=3 arrival in first minute|3 arrival in first 2 minute)

=P(K=3 in first and 0 in second minute)/P(3 arrival in first 2 minute)

=(e^-1/1!)*(e^-0/0!)/(e^-2(2)³/3!) =1/8

d)

here P(A U B) =1-0.2 =0.8

hence P(B) =P(A U B)+P(A n B)-P(A)=0.8+0.1-0.4=0.5