Question

A binomial random variable has \( n=5 \) and \( p=0.2 \) What is the probability of 3 successes or fewer? Express your answer rounded to four decimal places.

Answer 1

The probability of having 3 successes or fewer in a binomial random variable with n=5 and p=0.2 is approximately 0.9933, rounded to four decimal places.

Explanation:

To find the probability of 3 successes or fewer in a binomial random variable with n=5 and p=0.2, we can calculate the cumulative probability of getting 0, 1, 2, or 3 successes.

q=1-p=1-0.2=0.8

Using the binomial probability formula:

P(X
`\leq`3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

Where P(X=k) is the probability of getting exactly k successes in a binomial distribution.

P(X=0)=(5C0)*(0.2)^0*(0.8)^5-0=0.32768

P(X=1)=(5C1)*(0.2)^1*(0.8)^5-1=0.4096

P(X=2)=(5C2)*(0.2)^2*(0.8)^5-2=0.2048

P(X=3)=(5C3)*(0.2)^3*(0.8)^5-3=0.0512

Now, we can add these probabilities to find the cumulative probability:

P(X≤3)=0.32768+0.4096+0.2048+0.0512=0.9933

Therefore, the probability of having 3 successes or fewer in a binomial random variable with n=5 and p=0.2 is approximately 0.9933, rounded to four decimal places.