Question

Assume that a procedure yields a binomial distribution with a trial repeated n=5 times. Find the probability distribution given the probability p=0.533 of success on a single trial.

(Report answers accurate to 4 decimal places.)

k	P(X = k)
0
1
2
3
4
5

Answer 1

Here, n = 5, p = 0.533, (1 - p) = 0.467 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 5C0 * 0.533^0 * 0.467^5
P(X = 0) = 0.0222

P(X = 1) = 5C1 * 0.533^1 * 0.467^4
P(X = 1) = 0.1268

P(X = 2) = 5C2 * 0.533^2 * 0.467^3
P(X = 2) = 0.2893

P(X = 3) = 5C3 * 0.533^3 * 0.467^2
P(X = 3) = 0.3302

P(X = 4) = 5C4 * 0.533^4 * 0.467^1
P(X = 4) = 0.1884

P(X = 5) = 5C5 * 0.533^5 * 0.467^0
P(X = 5) = 0.0430

k	P(X = k)
0	0.0222
1	0.1268
2	0.2893
3	0.3302
4	0.1884
5	0.0430