Question
How do I plug this into the Binomial Probability Formula?
prob(lost) = 5/1000 = 1/200
prob(not lost) = 199/200
prob( 2 of 22 are lost)
= C(22,2) (1/200)^2 (199/200)^20
= .005224..
prob(lost) = 5/1000 = 1/200
prob(not lost) = 199/200
prob( 2 of 22 are lost)
= C(22,2) (1/200)^2 (199/200)^20
= .005224..