TS Contributor
a little help with this one will also be appreciated !

The number of bugs in a computer is a Poisson random variable with a rate parameter of 3 bugs a year. When a bug occurs, a technician A is called to fix it with probability 0.5, a technician B is called to fix it with probability 0.2 and a technician C with probability 0.1. The probability that C won't be called to fix any bug during a year is:

a. e^-2.7 b. e^-0.3 c. 1-e^-2.7 d. 1-e^-0.3

cheers !!


New Member
prob to have n bugs: exp(-3) 3^n/n!

prob not to call C provided n bugs occur: 0.9^n

total prob not to call C:

sum (over n) exp(-3) 3^n/n! times 0.9^n = exp(-3)*exp(3*0.9) = exp(-0.3)