Question

Suppose you just got onto a road, whose length is 10 miles. At each of the 10 mile-markers on the road (namely, at the end of mile 1, end of mile 2,. . . , end of mile 10 of the road), there is a 20% probability of there being a police car trying to catch speeders, independently for each mile. You proceed as follows. Until you see your first police officer, you drive at 60 miles per hour. After you see your first police officer, you drive the rest of the distance at 50 miles per hour. Let X be the random variable that denotes the number of miles you drive on this road until you see a police car; if you never see a police car on this road, then let X = 10 as the max road length is only 10 miles.

Express the amount of time it takes you to reach the end of the road as a function of X.
Write the Probability Mass Function of X (list all values that X can take and for each value x, give the expression for P(X = x))
What is the relationship between the distribution of X and the geometric distribution?

Answer 1

X =number of miles it Crosses without police X can take values 1,2,3,4,5,6,7,8,9,10 X = 1 means there is police at the end of 1 mile P(X = 1) = 0.2 P(X = 2) = 0.8 *0.2 P(X = 3) = 0.82 * 0.2 P(X = 4) = 0.8% * 0.2 P(X = 10) = 0.89 * 0.2 X = 3 means 1miles nopolice, 2mile no police, at 3 mile there is police. All these are independent so probability will be 0.8*0.8*0.2. similarly others. 10 to make this a pmf it should satisfy CP(X = k) = 1 k=1 C- C, PCX = k) [,0.341012 1 0.2 + 0.8 * 0.2 + 0.82 * 0.2 + 0.83 * 0.2 + ... + 0.89 * 0.2 (1 + 0.8 +0.82 +0.83 + ..... + 0.890.2 P(X = k) = (1 - 0,890.0 0.2(1 – 0.8%)/0.2 (1 – 0.89) so pmf is given by -0.8k-10.2 ;k = 1,2,3,4,5....10 X 10 - X amount of time it takes to reach the end of the road is an t- the relationship between this pmf and geometric distribution is that this pmf is the truncated pmf of geometric distribution