Question

There are three routes from a person's home to her place of work. There are four parking lots where she works, three entrances into her building, two elevators to her floor, and one route from each elevator to her office door. a) How many ways can she go from her home to her office? [2 marks] b) If she makes her various choices at random, what is the probability that she will take Morningside Drive, park in lot A, use the south entrance, and take elevator 1? [3 marks] c) As she starts her car one morning, she recalls parking lots A and B are closed for repair. What is the probability that she will take Industrial Avenue, park in lot D, use the north entrance, and take elevator 2?

Answer 1

a) 72

b) 1/72

c) 1/36

Explanation:

a) number of ways she can choose route= 3C1 = 3

number of ways she can choose parking lots= 4C1 = 4

number of ways she can choose entrances= 3C1 = 3

number of ways she can choose elevators= 2C1 = 2

number of ways she can go to office= number of ways she can choose route×number of ways she can choose parking lots×number of ways she can choose entrances×number of ways she can choose elevators

number of ways she can go to office= 3×4×3×2

= 72

b) Probability of morning side= number of morning side/ total number of routes= 1/3

probabiltiy of Parking lot A= number of parking lot A/ total number of parking lot= 1/4

probability of south entrance= number of south entrance/ total number of entrances= 1/3

probablity of elevator 1= number of elevator 1/ total number of elevator= 1/2

combine probability= 1/3× 1/4×1/3×1/2 = 1/72

c) Probability of Industrial Avenue= number of industrial avenue/ total number of avenue= 1/3

Probability of parking lot D= number of parking lot D/ total number of parking lot after deducting number of parking lots A and B = 1/2

Probability of north entrance= number of north entrance/ total number of entrance= 1/3

probablity of elevator 2= number of elevator 2/ total number of elevator

= 1/2

combine probability= 1/3 × 1/2 × 1/3 ×1/2

= 1/36