# Finding the probability that the motorist must stop at only one traffic light, at least two traffic lights [closed]

There are three traffic lights that a motorist must pass on the way to work. The probability that the motorist has to stop at the first traffic light is 0.2, and that for the second and third traffic lights are 0.5 and 0.8 respectively. Find the probability that the motorist has to stop at :

i) Only one traffic light

ii) At least two traffic lights

• Part (i) has been covered already. For part (ii); note that "at least two" is the same as 1-("stops at zero traffic lights"+"stops at one traffic light"). Apr 2, 2018 at 22:40
• P (At least 2 traffic lights) = 1 - ( 0 - 1.5) ?? Apr 2, 2018 at 23:00

For part i)

Lets list all the cases.

The motorist stops at the first traffic light but not at the other two.

The motorists stops at the second traffic light but not at the first and the third one

The motorists stops at the last traffic light but not at the first two.

What is the probability of any of the above happening?

• P (First traffic light) = 0.2/1.5 ... P (Second traffic light) = 0.5/1.5 ... P (3rd) = 0.8/1.5 .... 0.2 + 0.5 + 0.8 = 1.5 Apr 2, 2018 at 22:51
• No, that is incorrect. P(First traffic light)=$0.2\times(1-0.5)\times(1-0.8)$ , not $0.2/1.5$ Apr 3, 2018 at 15:37
• Okay, Therefore P (Second traffic light) = 0.5 × (1 - 0.2) × (1 - 0.8) and P (Third traffic light) = 0.8 × (1 - 0.2) × (1 - 0.5) ... Therefore P (First Traffic light) = 0.2 , P ( Second traffic light) = 0.08 and P ( Third traffic light) = 0.32 ... Correct? Apr 4, 2018 at 0:04
• Correct except its $0.02$ for the first one, double check that, and then to get the final answer you add them all up. Apr 4, 2018 at 17:21
• So the probability that the motorist has to drop at $only$ $one$ traffic light is 0.02 + 0.08 + 0.32 = $0.42$ ... what about the probability that the motorist has to drop at at least $TWO$ traffic lights? Apr 5, 2018 at 0:05