## nickersia 2 years ago John drives to work and passes three sets of traffic lights. The probability that he has to stop at the first is 0.6, second 0.7, third 0.8. He arrives late if he has to stop at any two sets of traffic lights. Calculate the probability that he is late. Answer is 0.788, how do I get it?

1. kropot72

P(stops at first & second) = 0.6 * 0.7 = 0.42 P(does not stop at second) = 1 - 0.7 = 0.3 P(stops at first & third) = 0.6 * 0.3 * 0.8 = 0.144 P(does not stop at first) = 1 - 0.6 = 0.4 P(stops at second and third) = 0.4 * 0.7 * 0.8 = 0.224 P(arrives late) = 0.42 + 0.144 + 0.224 = ?

2. nickersia

Oh! I understand now how it's done! I did it slightly different just to make it more clear for me :) P(does not stop at third) = 0.2 P(stops at first and second) = 0.6 * 0.7 * 0.2 = 0.084 P(does not stop at second) = 0.3 P(stops at first & third) = 0.6 * 0.3 * 0.8 = 0.144 P(does not stop at first) = 0.4 P(stops at second and third) = 0.4 * 0.7 * 0.8 = 0.224 P(stops at first, second and third) = 0.6 * 0.7 * 0.8 = 0.336 P(late) = 0.084 + 0.144 + 0.224 + 0.336 = 0.788 Thank you very much for all the help! I appreciate it! :)

3. kropot72

You're welcome :)