I have 3 independent variables at 20% each what is the percentage of success for 2 of them

1. Define the problem using binomial distribution: Each variable has a probability of 0.2 (20%) for success and 0.8 (80%) for failure. We are looking for the probability of exactly 2 successes out of 3 trials.

2. The general formula for the binomial probability is given by:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where:
- \( n \) is the total number of trials (3 in this case),
- \( k \) is the number of successes we want (2 in this case),
- \( p \) is the probability of success for each trial (0.2 in this case).

3. Substitute in the numbers:

P(X = 2) = \binom{3}{2} (0.2)^2 (0.8)^1

4. Calculate the binomial coefficient:

\binom{3}{2} = \frac{3!}{2!(3-2)!} = 3

5. Plug in the values:

P(X = 2) = 3 \times (0.2)^2 \times (0.8)^1

6. Simplify the expression:

P(X = 2) = 3 \times 0.04 \times 0.8

7. Complete the calculations:

P(X = 2) \approx 0.096

Therefore, the probability of success for exactly 2 out of 3 independent variables, each with a 20% success rate, is approximately 9.6%.