Exercise 2.55. Peter and Mary take turns throwing one dart at the dartboard. Pet
ID: 3045390 • Letter: E
Question
Exercise 2.55. Peter and Mary take turns throwing one dart at the dartboard. Peter hits the bullseye with probability p and Mary hits the bullseye with prob- ability r. Whoever hits the bullseye first wins. Suppose Peter throws the first dart. (a) What is the probability that Mary wins? b) Let X be the number of times Mary throws a dart in the game. Find the possible values and the probability mass function of X. Check that the function you give sums to 1. Is this a familiar named distribution? c) Find the conditional probability P(X = kl Mary wins), for all possible values k of X. Is this a familiar named distribution?Explanation / Answer
(a) Here Pr(Mary wins) = Pr(Peter loses the first game) * Pr(mary wins the game) + Pr(peter and mary both losses the first game) * Pr(peter loses his second chance) * Pr(mary win the second chance) + ....
= (1-p) * r + (1-p)2(1-r)* r + (1-p)3 (1-r)2r + .....
= (1-p)r [ 1 + (1-p) * (1-r) + ...]
= (1-p) r [1/ {1 - (1-p)(1-r)}]
= (1-p) r/ (r + p - rp)
(b) Let X is the number of times mary throws a dart in the game.
f(X) = (1-p)x(1-r)x-1 r + (1-p)x(1-r)x p = (1-p)x(1 - r)x-1 [r + (1-r)p] = (1-p)x(1 - r)x-1[r + p - rp]
THese two terms means that Peter failed in X times and mary win it in xth times plus both failed in Xth time but peter won in (x+1)th time.
The distribution is geometric one. THe distribuation functions gives sum to 1.
(c) P (X = k l mary wins) = (1-p)k(1 - r)k-1r /[(1-p) r/ (r + p - rp)] = (1-p)k-1(1-r)k-1 (r + p - rp)
it is also a geometric distribution with probability of success = r + p - rp
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