In tests of stopping distances for automobiles the automobil

In tests of stopping distances for automobiles, the automobiles
tended to stop at a distance uniformly distributed between a and b.

a) Find the probability that one of these automobiles
   stops closer to distance a than to distance b.

b) Suppose three automobiles are used in a test ... Find the
   probability that exactly one of the three travels past
   the midpoint between a and b.

Solution

since it is uniform distribution so pdf is given by

f(x) = 1/b-a

1.

probability that stops closer to a than b = stops between a and (b-a)/2.

P(a<X<(b-a)/2) = int_{a}^{b-a/2}1\\(b-a)dx

= 1/(b-a)(a-(b-a)/2)

=(3a-b)/2(b-a)

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2.

one passed the midpoint.while other two stops before midpoint.

so probability is = (3a-b)/(2(b-a))*(3a-b)/(2(b-a))*(b+a)/2(b-a)

=(b+a)(3a-b)^2/8(b-a)^3