Question

Bobby receives a gift for his tenth birthday — a model car. The model car has...

Bobby receives a gift for his tenth birthday — a model car. The model car has three compo- nents, which we call A, B, and C. Unfortunately, Bobby does not like the gift and consequently uses it for target practice. He throws rocks at the model car with the goal of destroying it as soon as possible. (When the car is finished, he will start destroying his other unsatisfactory gifts. Darling child.)

For any given throw, the probability that that the rock hits Component A is πA = 0.1, the probability that it hits Component B is πB = 0.2, and the probability that it hits Component C is πC = 0.3. These events are independent of one another and of the condition of the car.

For example, whether or not Component A is hit by the first rock, it will still be hit by the second rock with probability 0.1.

Once a component is hit by a rock, that component will stop working. The car will be broken once Component A stops working or bothComponents B and C stop working.

What is the probability that Bobby destroys the car on the Nth throw?

N=2

N=3

Homework Answers

Answer #1

P( rock does not hit any of the 3 components ) = 1 - (0.1 + 0.2 + 0.3) = 0.4

For the car to get destroyed in 2nd throw,

(i) First throw does not hit any of the components and 2nd throw hits A , p1 = 0.4*0.1 =0.04

(ii) First throw hits B and 2nd throw hits either A or C = 0.2*(0.1 + 0.3) = 0.08

(iii) First throw hits C and 2nd throw hits either A or B = 0.3*(0.1 + 0.2) = 0.09

Thus, total probability for the car to be destroyed in 2nd throw = 0.04 + 0.08 + 0.09 = 0.21

For the car to be destroyed in 3rd throw,

(i) First throw hits B, 2nd throw hits (B or none) and 3rd throw hits either A or C = 0.2*(0.4 + 0.2) * (0.1+ 0.3)= 0.048

(ii) First throw hits C, 2nd throw hits (C or none) and 3rd throw hits either A or B = 0.3*(0.4 + 0.3) * (0.1+ 0.2)= 0.063

(iii) First throw does not hit any of the components * Probabilty that the car is destroyed in 2nd throw = 0.4*0.21 = 0.084

Thus, required probability of the car to be destroyed in 3rd throw = 0.048 + 0.063 + 0.084 = 0.195

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