P = 0, never happens
P = 1, always happens
binomial probability
P(n,r) = C(n,r) m^r (1-m)^(n-r)
for exactly r = 1 for n = 2
C(2,1) = 2!/[1!(2-1)!] = 2
P(2,1)= 2 m^1 (1-m)^1 = 2 m(1-m)
in other word yes-no or no-yes :)
now for twice
P(2,2) = 1 m^2 (m-1)^0 = m^2
which we all know anyway
so for either
P(2,2)+P(2,1) = m^2+2m(1-m) =2m-m^2
Given P(A) = m is the probability of event A occurring in any given trial:
1. State the range of possible values of m
2. Suppose two trials are performed independently. Find, in terms of m, the probability of A occurring:
a) Exactly once
b) At least once
I am very confused of how to solve this because it only involves variables with no numbers.
1 answer
1 answer