That is not how a binomial probability is calculated. Consider the case where you flip a coin twice - do you have a 2 * 1/2, or 100% chance of getting a 'head'? Obviously not – you have a 75% chance.
The actual way you calculate the probability of an event with a binomial distribution is by taking the inverse probability and raising it to the power of the attempts, then subtracting from 1. So in the case of two coin flips, that is 1 - (1/2 ^ 2), or a 75% of getting a 'head'. That would mean, with a huge host of assumptions, that a motorcyclist's chance of dying over 30 years would be 1 - (99928/100000)^30, or (funnily enough), 2.13%.
By the binomial series [1], the true value is 1-(1-p)^n = np - n(n-1)/2! p^2 + n(n-1)(n-2)/3! p^3 + ...
So approximating it to just np corresponds to taking just the leading term of the series. Of course, this is only valid when p is small enough. To see how good the approximation is, we can compare it to the two first terms of the series, which is np - n(n-1)/2 p^2, or approximately np - (np)^2 /2.
So the approximation np is about twice the percentage difference between the approximation and the true value. In this case, the approximation np is about 2%, and that guess is itself about 1% wrong.
You set out to correct a naive statistical assumption, computed it properly, and the result was about the same (not at all common for such problems). But you posted it anyway, in the spirit of sharing useful information.
The actual way you calculate the probability of an event with a binomial distribution is by taking the inverse probability and raising it to the power of the attempts, then subtracting from 1. So in the case of two coin flips, that is 1 - (1/2 ^ 2), or a 75% of getting a 'head'. That would mean, with a huge host of assumptions, that a motorcyclist's chance of dying over 30 years would be 1 - (99928/100000)^30, or (funnily enough), 2.13%.
edit: I said bimodal when I meant binomial.