The P(E) = 2/5. What is P(E')?
2007-06-04 04:57:01 · 4 answers · asked by tom m 1 in Science & Mathematics ➔ Mathematics
P(E') asks for the possibility of an event that is not E.
The largest probability of an even happening is 1, and so the probability of P(E) = 2/5 means that the probability of E not happening is =1-2/5
therefore, P(E') = 3/5
2007-06-04 05:04:50 · answer #1 · answered by shekum 2 · 0⤊ 0⤋
You don't have enough information to solve. I'm not even sure what you are trying to represent. It looks like P(E) is a function of E, but I don't know what P(E') means. P'(E) would be the first derivative of P, and we'd assume it is with respect to an independent variable E. If E is really a function E(x), the derivative would be with respect to x, but you'd need to make that clear. In any event, the value of a function at one point is not enough to determine the first derivative at that point.
If P(E) = 2/5 is a constant function, then P'(E) = 0. P(E') is not meaningful.
2007-06-04 11:59:31 · answer #2 · answered by DavidK93 7 · 0⤊ 0⤋
P(E) is probability of event E happening= 2/5
P(E`) is probability of E not happening = 3/5
2007-06-04 12:14:45 · answer #3 · answered by Como 7 · 0⤊ 0⤋
zero
2007-06-04 11:59:47 · answer #4 · answered by Anonymous · 0⤊ 0⤋