Answer:
0.264 ; 0.079
Explanation:
Given that:
Sample size, n = 100
Probability of being active, p = 1% = 1/100 = 0.01
Using the binomial probability relation :
P(x =x) = nCx * p^x * (1 - p)^(n - x)
Probability that more than 1 user will be active
P(x > 1) = 1 - [p(x=0) + p(x = 1)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x > 1) = 1 - [0.366 + 0.370]
P(x > 1) = 0.264
2.)
Probability that more than 2 user will be active
P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185
P(x > 1) = 1 - [0.366 + 0.370 + 0.185]
P(x > 1) = 0.079
