A shooter hits a target with probability 0.4 on each shot, independently of the other shots…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A shooter hits a target with probability on each shot, independently of the other shots. She keeps shooting until she hits the target for the first time, but she stops after shots whether or not she has hit it. Let be the number of shots she takes. Find the probability that she needs more than shots, which means that the first six shots all miss. Find the expected number of shots . Find the probability that her first hit is on the third shot. The range officer wants the expected number of shots to plan the supply of cartridges. Give decimals to four places for the first two answers, and to three places for the third. Think about how the geometric distribution changes when the number of shots is capped.
Need a reminder? The key ideaShow
, and . For large , , and the continuity correction uses for .
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Six misses. .
Step 2. Expected shots. The number of shots is at least exactly when the first shots miss, so . Then .
Step 3. First hit on the third shot. Two misses and then a hit: .
Answer: , about and .
Compare. With no cap, the expected number of shots until the first hit would be . The cap at shots lowers it only to , because most of the time she hits long before the sixth shot.
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