ONLY ANSWER QUESTION D 1. (4 pts each) Ivan has a Hot Wheels track upon which he

Question

ONLY ANSWER QUESTION D

1. (4 pts each) Ivan has a Hot Wheels track upon which he runs his Hot Wheels, a collection 74 small toy cars. The track has a single lane, so if the first car on the track does not go very far then the second car run on the track is likely to collide into the first, causing gales of laughter from Ivan. We will call this event a crash. He rarely clears the track before running yet another car, so a third car may in fact run into the two car pile-up causing a three car answer the following pile-up. In this case we say all three cars were involved in a crash. Please about Ivan's Hot Wheel cars and the number of crashes (a) Ivan picks 6 Hot Wheel cars from his collection to run upon the track. How many different selections are possible? (b) Ivan lines up his 6 selected cars to run upon the track (i.e. puts them in an order). How many different line ups are possible? (c) One of the cars Ivan selected was the batmobile. This car is heavier than the rest and has straight, easy-rolling wheels implying it will travel farther on the track than any of the other cars (say with 100% probability). If Ivan does not remove cars from the track. what are the chances that the batmobile will crash into another vehicle? Here you may assume that all possible line-ups are equally likely (and that was a hint) (d) Another of the six cars is the blue pizza delivery van which is missing one wheel (too many crashes I guess). This car is the slowest in the collection and will always travel the shortest distance on the track. Calculate the expected number of times this car is involved in a crash. e) Suppose you know that the batmobile is second in the line-up. How does this new information change the expected number of crashes involving the pizza delivery van?

Explanation / Answer

We know that the blue pizza delivery van is the slowest of the cars. Let us call it BPDV.

We also know that the Bat-mobile (BM) is another of the 6 cars.

One point of information is NOT mentioned in the question. Does the fact that the Bat-mobile is the heaviest of the lot mean that it carries with it, all the cars it crashes into, to it's original length of free travel?

i) If 'NO':-

If the BPDV is the i-th car, all the cars coming afterwards will crash into it and it will NOT crash into any cars coming before it.

Hence, the number of times BPDV is involved in a car-crash is equal to (6-i), where i = position of BPDV in line-up.

Let 'number of crashes BPDV is involved in' be denoted by X

=> E[X] = Sum of [ (Probability of line-up position 'i')*(Number of crashes when in position 'i') ]

= Sum of [1/6*(6-i)], where i = 1 to 6 = 15/6 = 5/2 = 2.5 [Answer]

ii) If 'YES':-

If the BPDV is in the i-th position and the BM comes at j-th position which is AFTER the BPDV, then there will only be: (6-i)-(6-j) crashes = j-i (j > i)

If the BPDV is in the i-th position and the BM comes at j-th position which is BEFORE the BPDV, then there will be: (6-i) crashes (j < i)

Probability of each (i,j) pair = (1/6)*(1/5) = 1/30 (6C2 ways to pick any pair of positions)

Now, each pair [say: (3,1)] will have two cases: one with i > j (3,1) and the other with i < j (1,3)

Let 'number of crashes BPDV is involved in' be denoted by X

E[X] = [ Sum of (1/30)*(j-i) for j = 2 to 6 and i < j ] + [ Sum of {(1/30)*(6-i)} for i = 2 to 6 and j < i ]

= (1/30)*(1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 3 + 4 + 1 + 2 + 3 + 4 + 5) + (1/30)*(4 + 3 + 2 + 1 + 0)

= (1/30)*(35+10) = 45/30 = 1.5 [Answer]

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