Answer :
Answer:
a) 0.05
b) 0.9826
c) 0.000039308
Step-by-step explanation:
a) [tex] P_X(0) = \frac{e^{-3}3^0}{0!} = e^{-3} = 0.05[/tex]
b) For two minutes, the mean is doubled, hence it is 6. In order to calculate the probability of al least two calls arriving, we calculate first the probability of the complementary event: At most 1 call will arrive. For that probability, we need to sum the probabilities of 0 and 1.
[tex]P_X(0) = e^{-6} [/tex]
[tex]P_X(1) = \frac{e^{-6}*6^1}{1!} = 6*e^{-6}[/tex]
Hence,
[tex] P(X \geq 2) = 1-P(X < 2) = 1- 7*e^{-6} = 0.9826[/tex]
c) For five minutes the mean is 15. We need to sum the probabilities of 0, 1 and 2.
[tex] P_X(0) = e^{-15}[/tex]
[tex] P_X(1) = e^{15}*15 [/tex]
[tex]P_X(2) = \frac{e^{-15}*15^2}{2!} = 112.5*e^{-15} [/tex]
As a result,
[tex] P(X \leq 2) = e^{-15}(1+15+112.5) = 0.000039308 [/tex]
Practically 0