Kids arrive on a playground according to a Poisson distribut

Kids arrive on a playground according to a Poisson distribution, with arrival rate l = 1 every 10 minutes. They would like to play a game that requires at least three players.

What is the probability that in the first ½ hour they will have enough players to play the game?

Assuming that there are two kids already on the playground. What is the probability that they would have to wait at least 10 more minutes to get the third player? (Hint: use the distribution for the inter-arrival times.)

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials
a)
arrival rate l = 1 every 10 minutes
For 30 mins the mean will be 3

P(at least three players) = P( X > = 3 ) = 1 - P (X < 3)

P( X < 3) = P(X=2) + P(X=1) + P(X=0) +   
= e^-3 * 0 ^ 2 / 2! + e^-3 * ^ 1 / 1! + e^-3 * ^ 0 / 0! +
= 0.4232

P( X > = 3 ) = 1 - P (X < 3) = 0.5768
b)
there are two kids already on the playground
For next 10 mins there is no kid arrived and next min 1 kid arrived
P( Atleast 10 mins waited) = P( X = 0 ) + P( X = 1 ) = e ^-1 * 1^0 / 0! + e ^-1 * 1^1 / 1! = 0.3679 + 0.3679 = 0.7358