# [Math] Motor vehicle at an intersection probability

probabilityprobability distributions

Motor vehicles arriving at an intersection can turn right or left or continue
straight ahead. In a study of traffic patterns at this intersection over a long
period of time, engineers have noted that 40 percents of the motor vehicles
turn left, 25 percents turn right, and the remainder continue straight ahead.
For the next five cars entering the intersection, what is the probability that
at least one turn right?

Let $L$ be the event that the vehicle turns left, $S$ be the event that the vehicle goes straight, and $R$ the event the vehicle turns right. Thus, the probability that the vehicle turns left OR goes straight is $P(L) + P(S)$ since the events are mutually exclusive and independent. Call this probability $P(LS)$. Then $P(LS)$ = $P(L)$ + $P(S)$ = 0.40 + 0.35 = 0.75.
Given the above, the probability that the next 5 cars go straight or left is equal to $P(LS)^5$ due to independence. So, the probability that at least 1 car goes right is simply the complement of this; that is, $P(R) = 1 - P(LS)^5$ = 1 - $0.75^5$ = 0.7626953.