Question

I need to find at most 3 successes x=0, x=1, x=2,x=3 n=20 p=0.05 need to solve all of them and add together

Answer 1

The binomial distribution is used to calculate the probability of repeated successes if we know the individual odds of success of each event.

The formula is:

`P(n,k)=\binom{n}{k}p^kq^(n-k)`

Where n is the number of trials, k is the number of successes, p is the individual success probability, and q = 1 - p.

For n = 20, p = 0.05, it's required to find:

P(x ≤ 3) = P(20, 0) + P(20, 1) + P(20, 2) + P(20, 3)

Applying the formula, for q = 1 - p = 0.95

`P(20,0)=\binom{20}{0}0.05^00.95^(20-0)`

Operating:

`P(20,0)=\frac{20!}{0!\text{ }20!}1\cdot0.3585=0.3585`

Calculate:

`P(20,1)=\binom{20}{1}0.05^10.95^(19)`

Operate:

`P(20,1)=\frac{20!}{1!\text{ 19}!}0.05\cdot0.3774=0.3774`

Calculate:

`P(20,2)=\binom{20}{2}0.05^20.95^(18)`

Operate:

`P(20,2)=\frac{20!}{2!\text{ 18}!}0.0025\cdot0.3972=190\cdot0.0025\cdot0.3972=0.1887`

Calculate:

`P(20,3)=\binom{20}{3}0.05^30.95^(17)`

Operate:

`P(20,3)=\frac{20!}{3!\text{ 17}!}0.000125\cdot0.4181=1140\cdot0.000125\cdot0.4181=0.0596`

The total probability is:

P(x ≤ 3) = 0.3585 + 0.3774 + 0.1887 + 0.0596 = 0.9842

P(x ≤ 3) = 0.9842