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candy1400 · May 26, 2011, 10:29 PM

The probability is 1 in 4,000,000 that a single auto trip in the United States will result in a fatality. Over a lifetime, an average U.S. driver takes 50,000 trips. What is the probability of a fatal accident over a lifetime? Why might the assumption of independence be violated?

ebaines · May 27, 2011, 06:40 AM

Think through it this way:

If the probability of something occurring is $p$ , then the probability of that something not occurring is $1-p$ . The probability of that something occurring at least once in $n$ trips is calculated by subtracting from 1 the probability that that something does not occur in n tries:

prob(occur at least once) = 1- prob(never occurs) = $1 - (1-p)^n$

This formula can be expanded like this:

$<br /> <br /> 1 - (1-p)^n = 1 - (1 - n p + C(n,2) p^{n-1} -C(n,3)p^{n-2} + ... -n p^{n-1} + p^n) \\<br /> = n p - C(n,2) p^{n-1} +C(n,3)p^{n-2} - ... +n p^{n-1} - p^n)<br />$

For small values of $p$ this is very close to the value of $np$ .

Here's an example: suppose there is a 1 in 100 probability that a baseball player will hit a home run. What is the probability of hitting at least one home run in 10 at bats? Well, the probability of not hitting a home run in any one at bat is 99/100. The probability of not hitting a home run in 10 at bats is (99/100)^10. So the probability of hitting at least one home run in 10 at bats is 1 - (99/100)^10. This turns out to be 0.096, which is fairly close to $np=10 \times 0.1 = 0.1$ .

Can you now apply this technique to your homework problem? Post back with what you get for an answer.