Cars can either turn either left or right Suppose that succe

Cars can either turn either left or right. Suppose that successive cars choose a turning direction independently of one another and that P(turning left) = 0.6

Among the next 10 cars, what is the probability that 3 cars turn left?

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

probability that 3 cars turn left = P( X = 3 ) = ( 10 3 ) * ( 0.6^3) * ( 1 - 0.6 )^7
= 0.0425