P(all) = 3C3 (.67)^3(.33)^0 = 300763
Answer is 0.301
You take a trip by air that involves three independent flights. If there is an 67% chance each specific leg of the trip is on time, what is the probability all three flights arrive on time? (Round your answer to 3 decimal places.)
Probability =??
2 answers
Typo
P(all) = 3C3 (.67)^3(.33)^0 =0. 300763
Answer is 0.301
P(all) = 3C3 (.67)^3(.33)^0 =0. 300763
Answer is 0.301