Using the Binomial distribution, If \( n=10 \) and \( p=0.6 \), find \( P(x=4) \)

We are given a binomial distribution problem with \( n=10 \) trials and probability of success \( p=0.6 \). We want to find the probability that exactly \( x=4 \) successes occur, using the formula for the binomial probability: \[ P(x=k) = \binom{n}{k} p^k (1-p)^{n-k} \] For \( k=4 \), the formula becomes: \[ P(x=4) = \binom{10}{4} (0.6)^4 (0.4)^6 \] **Step 1: Compute the binomial coefficient** \[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210 \] **Step 2: Compute the powers** \[ (0.6)^4 = 0.6 \times 0.6 \times 0.6 \times 0.6 \approx 0.1296 \] \[ (0.4)^6 = 0.4 \times 0.4 \times 0.4 \times 0.4 \times 0.4 \times 0.4 \approx 0.004096 \] **Step 3: Combine all components** \[ P(x=4) = 210 \times 0.1296 \times 0.004096 \approx 210 \times 0.0005314416 \approx 0.1116 \] Thus, the probability that exactly 4 successes occur is: \[ P(x=4) \approx 0.1116 \]