The probability is 1 in 4,000,000 that a single auto trip in the United States will resultin a fatality. Over?

Question

The probability is 1 in 4,000,000 that a single auto trip in the United States will resultin 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 carfully. Hint: Assume independent events. Why migth the assumtion of independence be violated? (B) Why might a driver be tempted not to use a seat belt "just on this trip"?

Answer 1

Let X be the number of fatal  accidents a drive could be in during their life time.  X has the binomial distribution with n = 50000 trials and success probability p = 2.5e-07



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 50000 , p = 2.5e-07 )

the mean of the binomial distribution is n * p = 0.0125

the variance of the binomial distribution is n * p * (1 - p) = 0.01250000

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 0.1118034

P( X =  0 ) =  0.9875778

P( X ≥ 1 ) = 1 - P(X = 0) = 1 - 0.9875778 = 0.0124222

for any one trip the probability of a fatality is 1/4000 000, so not using  seat belt for "just this one trip" seems somewhat reasonable because it is so unlikely.