According to an automobile manufacturer, the company uses 3,000 lock-and-key combinations on it's vehicles. Suppose that you find a key for one of those cars.
a) What is the expected number of vehicles that you would have to check to find one that fits your key?
b) What is the probability that you would have to check at least 3,000 vehicles to find one that your key fit?
Here are the answers to your questions. I'll start with question (c), as I feel it's important to know first what distribution we'll be using.
c) The probability distribution we'll use here is the geometric distribution. The geometric distribution shows the probability of needing n trials to succeed in an experiment that has equal probability of success in each trial.
In this case, we have a key that opens 1 in 3,000 cars. So, each time we try it on a car, we have a 1/3000 probability of succeeding (let's call this probability 'p'). This probability clearly remains constant as we keep trying the key on other cars.
The formula is derived in the following way. In order to need n trials to succeed, we must fail n-1 consecutive times. The ...
Expected Values and Probability Distributions are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.