In tests of stopping distance for automobiles, cars traveling 30 miles per hour

Question

In tests of stopping distance for automobiles, cars traveling 30 miles per hour before the brakes were applied tended to travel distances that appeared to be uniformly distributed between two points a and b . Find the probabilities of the following events. a One of these automobiles, selected at random, stops closer to a than to b . b One of these automobiles, selected at random, stops at a point where the distance to a is more than three times the distance to b . c Suppose that three automobiles are used in the test. Find the probability that exactly one of the three travels past the midpoint between a and b

Explanation / Answer

Given that distance of stopping is Uniformly distributed between a and b

we know that P(distance < d) = (d-a)/(b-a)

a)
probability that a automobile selected at random stops closer to a than to b = probability that a automobile selected at random stops before midpoint i.e (a+b)/2

P(disatnce < (a+b)/2) = ((a+b)/2 - a)/(b-a)

= (a+b -2a)/(2(b-a))

= (b-a)/(2(b-a))

= 1/2 = 0.5

b)

let k be the point at which distance from a is three times the distance from b => k-a = 3*(b-k)

=> k -a = 3b - 3k => 3b + a = 4k => k = (3b +a)/4

probability that atumobile selected at random stops at a point where the distance to a is more than three times the distance to b = probability that atumobile selected at random stops at a point after (3b+a)/4

P(distance > (3b+a)/4) = P(distance < (3+a)/4) = ((3b+a)/4 - a)/(b-a)

= (3b+a -4a)/(4(b-a)) = (3*(b-a))/(4*(b-a)) = 3/4

c)

probability that a automobile selected at random travels past the midpoint between a and b = 0.5

probability that exactly one of the three travels past the midpoint between a and b from three automobiles =

= 3*(0.5)1(0.5)2 = 3/8 = 0.375

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