Answer:
0.1302 or 13.02%
Step-by-step explanation:
To find the probability of exactly 2 out of 6 being brown, we can use the binomial probability formula:
\[ P(X=k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \]
Where:
- \( n \) is the number of trials (in this case, 6)
- \( k \) is the number of successes (in this case, 2)
- \( p \) is the probability of success on each trial (in this case, 12% or 0.12)
- \( \binom{n}{k} \) represents the number of combinations of \( n \) things taken \( k \) at a time, calculated as \( \frac{n!}{k!(n-k)!} \)
So, substituting the values:
\[ P(X=2) = \binom{6}{2} \times (0.12)^2 \times (1-0.12)^{6-2} \]
\[ P(X=2) = \binom{6}{2} \times 0.12^2 \times 0.88^4 \]
Now, let's calculate each part:
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
\[ (0.12)^2 = 0.0144 \]
\[ (0.88)^4 \approx 0.5997 \]
Now, multiply these values together:
\[ P(X=2) = 15 \times 0.0144 \times 0.5997 \]
\[ P(X=2) \approx 0.1302 \]
So, the probability that exactly 2 out of 6 will be brown is approximately 0.1302 or 13.02%.
