Learning programming and trying to understand this example.
Given multiple independent events, each with a probability of occurring, what is the probability of just one event occurring?
If we have 2 cases I can understand it this way:
P(exactly one event occurs) = P(a and not b) + P(not a and b)
Calculated as:
(Pa)*(1-Pb) + (Pb)*(1-Pa)
Can it be explained why we use (1-probability) here? What does subtracting the prob. from 1 "do"?
Thank you.
Generalized as:
n
∑ pi (1 − p1 )…(1 − pi−1 )(1 − pi+1 )…(1 − pn )
i=1
Coded in R as:
exactlyone <- function(p) {
notp <- 1 - p
tot <- 0.0
for (i in 1:length(p))
tot <- tot + p[i] * prod(notp[-i])
return(tot)
}
Best Answer
In general, if the probability of an event $A$ occurring is $$P(A)$$
then the probability of the event not occurring is $$1-P(A)$$
This is because either $A$ occurs, or it doesn't.
Another way of phrasing this is if $A'$ is "$A$ doesn't happen", then $$P(A) + P(A') = 1$$