Answer:
Step-by-step explanation:
Given that a parking lot has two entrances. Cars arrive at entrance I according to a Poisson distribution at an average of 3 per hour and at entrance II according to a Poisson distribution at an average of 2 per hour.
Assuming the number of cars arriving at the two parking lots are independent we have total number of cars arriving X is Poisson with parameter 3+2 = 5
X is Poisson with mean = 5
the probability that a total of 3 cars will arrive at the parking lot in a given hour
= P(X=3) = 0.1404
b) the probability that less than 3 cars will arrive at the parking lot in a given hour
= P(X<3)
= P(0)+P(1)+P(2)
= 0.1247
Write an expression for 213 increased by d: 
Answer:
(9sqrt13)/13
Step-by-step explanation:
0.21951 Reapeating cause the decimal does not terminate
We need to correctly choose exactly 4 out of the 6 drawn numbers.
Apply hypergeometric distribution:
a=number of correctly chosen numbers = 4
A=number of correct (drawn) numbers = 6
b=number of incorrectly chosen numbers = 2
B=number of undrawn numbers = 44-6 = 38
Then by the hypergeometric distribution
P(a,b,A,B)
=C(A,a)C(B,b)/C(A+B,a+b) [C(n,r)=combination of r objects taken out of n]
=C(6,4)C(38,2)/C(44,6)
=15*703/7059052
= 10545/7059052
= 0.001494 (to the nearest millionth)
Answer: probability of winning third prize is 10545/7059052=0.001494
