A fly disturbs Jack and Jills' happiness. Jack heroically attempts to swat it and let S be the number...?

Question

of attempts required by Jack to kill the fly. Suppose S has the following p.m.f.

P(S=s): .01, .05, .2, .74

As in bad comedy, for every attempt that misses the fly, Jack hits Jill with probability .21. What is the probability that Jack hits Jill at least once before killing the fly?

Answer 1

Dear ianboen86,

You haven't completely defined the probability mass function (p.m.f.) for S, but I am guessing you meant for the p.m.f. to take on the stated values for s = 1, 2, 3, and 4, respectively, and that the p.m.f. equals 0 otherwise.

Observe that the fly will be killed with certainty on the fourth swat if it was not killed on an earlier swat. This means that there will be at most three misses. Also observe that the probability that Jack hits Jill at least once is equal to 1 minus the probability that Jack misses Jill entirely. Because there are at most three misses swatting at the fly, this also means there are at most three misses of Jill by Jack as he is attempting to swat the fly. If we know the number of times Jack missed the fly, then we can compute the (conditional) probability that Jack misses Jill entirely.

Let f(y) be the probability that the fly is killed with exactly y misses. Thus,

f(0) = 0.01,
f(1) = 0.05,
f(2) = 0.20,
f(3) = 0.74, and
f(y) = 0 for all y not in {0, 1, 2, 3}.

Notice that f is essentially the same as the original p.m.f., except f is defined in terms of misses instead of swats, and the number of misses is always one fewer than the number of swats.

Next, let g(z) be the probability that Jack misses Jill entirely, given exactly z misses of the fly. We know that with each miss of the fly, Jill has a 0.21 probability of being hit. If we call that probability q, then the probability that Jill is missed with any particular miss of the fly is 1 - q = 0.79, which we will call r. Thus,

g(0) = r^0 = 0.79^0 = 1,
g(1) = r^1 = 0.79^1 = 0.79,
g(2) = r^2 = 0.79^2 = 0.6241,
g(3) = r^3 = 0.79^3 = 0.493039, and
g(z) = 0 for all z not in {0, 1, 2, 3}.

The probability that Jill is missed entirely and the fly is missed exactly m times is now determined by f(m)g(m). Summing these probabilities over all values of m then gives the probability that Jill is missed entirely, which we call e.

e = f(0)g(0) + f(1)g(1) + f(2)g(2) + f(3)g(3)
= (0.01)r^0 + (0.05)r^1 + (0.20)r^2 + (0.74)r^3
= (0.01)0.79^0 + (0.05)0.79^1 + (0.20)0.79^2 + (0.74)0.79^3
= 0.53917 (to five decimal places).

Letting h be the probability that Jack hits Jill at least once before killing the fly, we obtain

h = 1 - e
= 0.46083 (to five decimal places).