Ask Me Help Desk - Conditional Probability?

I could really use some help understanding the following problem:

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. (a) What is the probability of a fatal accident over a lifetime? Explain your reasoning carefully. Hint: Assume independent events. Why might the assumption of independence be violated? (b) Why might a driver be tempted not to use a seat belt “just on this trip”?

I was thinking conditional probability (given 1 in 4million fatalities what is probability that the average operator will have a fatal accident) but every time I calculate I end up with the number I started with 1 over 4,000,000.

A is event of fatality in a single auto trip
B is event that an average driver will have fatal accident

P(A) = .00000025, P(B) = .0125
not given the intersection

so, P(AintersectB) = P(A)xP(B)

P(A|B) = P(AnB)/P(B)

I can't tell you how to solve it (what I tried just freaked my calculator out), but I can tell you this isn't conditional. What's the condition? Think about what you wrote: you're saying what's the probability that someone will have a fatal accident on one particular trip given that they've already had a fatal accident at some point in their lifetime. That doesn't make sense. A condition means that you already know that some certain condition already exists. (e.g. what's the probability of the accident given they're driving a motorcycle.)

If "event that an average driver will have fatal accident" means that they will have the accident at some point in their lifetime, then B is the final answer you're looking for.

Since n is large and p is very small, use the Poisson distribution.

Let X be the number of accidents in a lifetime of trips, with parameter $\lambda=np$

$P(X=x)=e^{-\lambda}\frac{{\lambda}^{x}}{x!}$

You might want to read up more on how this distribution arises.

There's an easier way to get a very good estimate of the probability of being in an accident over 50,000 truips. Consider that the probability of NOT being in an accident for any single trip is 1 - 1/4,000,000. So the probability of not being in an accident for 50,000 trips in a row - assuming each trip is independent - is:

(1- 1/4,000,000)^50,000

You can't run this on a calculator as Morgaine300 found out, but you can take advantage of the polynomial expansion to make a very accurate estimate. Recall that:

$ (a-b)^n = a^n - na^{n-1}b +\frac {n(n-1)} 2 a^{n-2} b^2 - \frac {n(n-1)(n-2)} {3 \cdot 2} a^{n-3} b^3 +... $

For a = 1 you get:
$ (1-b)^n = 1 - nb + \frac {n(n-1)} 2 b^2 - \frac {n(n-1)(n-2)} {3 \cdot 2} b^3 + ...$

For small values of b you can truncate this after the first two terms, since terms with b^2, b^3 etc are all insignificant:

$ (1-b)^n = 1 - n*b $

So in this problem you have:

$ (1- \frac 1 {4,000,000} )^{50,000} = 1 - 50,000*\frac 1 {4,000,000} = 0.9875 $

This is the probability of not being in an accident 50,000 times in a row. Hence the probability of being in at least one accident is:

1- 0.09875 = 0.0125.