Assuming X is a binomial random variable, what is the probability exactly 3 cars are left unlocked?

Question

There is a p=12.5% chance a parked car will have been left unlocked. X is the number of cars left unlocked in a parking lot with a total of 6 cars. Assuming X is a binomial random variable, what is the probability exactly 3 cars are left unlocked?

P(X=3)=

Answer 1

P(X=3) = ₆C₃•(0.125³)•(0.875³) representing the number of ways that 3 of the 6 cars are unlocked and p, probability of unlocking, raised to power 3 along with 3 locked cars (individual probability 1-p).