I have forgotten my probability theory...

[elfs]

Oh, might LJ Brain, help me with this mathematics problem: a man is a demolition derbyist. The probability of him suffering an injury during any one derby is 0.12.

I know that the probability of him suffering an injury in two derbies is ((0.12 + 0.12) - (0.12 * 0.12)). What's the (generalized) formula for him suffering an injury over n derbies?

Comments

[alexrunningwolf.livejournal.com]

the formula you have above should be expandable to any number of Derbies, where P = Probabbility and N + the number of derbies.

the formula would look like ((N*P) - (P)to the power of N)

Hope that's wht you were asking for.

Wolf

[lovingboth]

Ambiguous question - do you mean 'he suffers at least one injury in n derbies' or 'he suffers an injury in each'?

The former = one minus the chance he isn't injured: 1 - (1 - 0.12) ^ n

The latter is simply (0.12) ^ n

[zanfur.livejournal.com]

Seconded. This is the formula you're looking for.

[whipartist.livejournal.com]

Yes, precisely. If you want to know what his chances of being injured at least once in N derbies, you really need to find the probability that he will not be injured in any of them, then subtract it from 1.

[neowolf2.livejournal.com]

Now suppose the chance he is injured in any particular race is 1/N, and he is in N races. As N becomes large, what is the chance he will be injured?

[blaisepascal]

I get 1-(1/e), or about 63%. You?

[neowolf2.livejournal.com]

That is correct!

[blaisepascal]

lovingboth has it, as far as I can tell. 1-(0.12)^n

[rkda.livejournal.com]

The probability that he will suffer an injury in one event is x = 0.12
The probability that he will NOT suffer an injury in one event is (1 - x) = (1.0 - 0.12) = .88
The probability that he will NOT suffer an injury in any of n events is .88^n or (1 - x)^n
So, the probability that he will be injured in n events is 1 - ((1 - x)^n)

[gchpaco.livejournal.com]

You want the binomial distribution. N trials (derbys), probability p of "success" 0.12, probability of zero successes in N trials is then (1 - p)^n and so the probability of him being injured is 1 - (1 - p)^n.

[srmalloy.livejournal.com]

And it was this math that made for a depressing statistic for bomber crews during WWII; the probability of making it back from any single mission was high, but over the twenty-five missions that you had to complete before being rotated home, your overall chance of completing your tour was pretty poor. The prevailing sentiment in 1943, when bombers were being lost at a high rate due to the lack of a fighter with the range to accompany them all the way to target and back (before the P-51), was "Who's gonna make twenty-five missions? We'll get shot down first."

[lovingboth]

The chance of coming back from any single mission varied with experience, of course, making the maths somewhat harder.