Question

the probability that a mature hen will lay an egg on a given day is 0.80. hannah has 12 hens. what is the probability that 10 of the 12 hens will lay eggs on a given day?
0.05
0.11
0.20
0.28

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \), where \( n = 12 \) (number of hens), \( k = 10 \) (number of hens laying eggs), \( p = 0.80 \) (probability a hen lays an egg).

Step2: Calculate the combination

First, calculate the combination \( \binom{12}{10} \). The formula for combinations is \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \). So, \( \binom{12}{10} = \binom{12}{2} = \frac{12!}{2!(12 - 2)!} = \frac{12 \times 11}{2 \times 1} = 66 \).

Step3: Calculate the probability

Now, substitute into the binomial formula: \( P(X = 10) = \binom{12}{10} (0.80)^{10} (1 - 0.80)^{12 - 10} \).

First, calculate \( (0.80)^{10} \approx 0.1073741824 \), \( (0.20)^{2} = 0.04 \).

Then, multiply them together with the combination: \( 66 \times 0.1073741824 \times 0.04 \approx 66 \times 0.0042949673 \approx 0.28346784 \approx 0.28 \)? Wait, no, wait, let's recalculate. Wait, maybe I made a mistake. Wait, \( (0.8)^{10} \) is approximately 0.10737, \( (0.2)^2 = 0.04 \). Then \( 66 * 0.10737 * 0.04 = 66 * 0.0042948 = 0.2834568 \approx 0.28 \)? But wait, maybe my calculation is wrong. Wait, no, let's check again. Wait, \( \binom{12}{10} = 66 \), \( p = 0.8 \), \( 1 - p = 0.2 \), \( k = 10 \), \( n = 12 \). So \( P = 66 (0.8)^{10} (0.2)^2 \). Let's compute \( (0.8)^{10} \): 0.8^2 = 0.64, 0.8^4 = 0.64^2 = 0.4096, 0.8^8 = 0.4096^2 = 0.16777216, 0.8^10 = 0.8^8 0.8^2 = 0.16777216 0.64 = 0.1073741824. Then \( (0.2)^2 = 0.04 \). Then 66 0.1073741824 = 7.0866960384. Then 7.0866960384 0.04 = 0.2834678415. So approximately 0.28. But wait, the options include 0.28. But wait, maybe I made a mistake. Wait, no, let's check with a calculator. Alternatively, maybe the correct answer is 0.28? Wait, but let's check the options. The options are 0.05, 0.11, 0.20, 0.28. So 0.28 is an option. Wait, but maybe my calculation is correct. Wait, but let's see, maybe I messed up the combination. Wait, \( \binom{12}{10} = \frac{12!}{10!2!} = \frac{12*11}{2*1} = 66 \), that's correct. Then \( (0.8)^{10} \approx 0.10737 \), \( (0.2)^2 = 0.04 \). Then 66 0.10737 0.04 = 0.283, which is approximately 0.28. So the answer should be 0.28.

Wait, but wait, maybe I made a mistake. Wait, let's check with another approach. Let's use the binomial probability formula correctly. So \( P(X=10) = C(12,10) (0.8)^10 (0.2)^2 \). Calculating \( C(12,10) = 66 \), \( (0.8)^10 = 0.1073741824 \), \( (0.2)^2 = 0.04 \). Then 66 0.1073741824 = 7.0866960384. Then 7.0866960384 0.04 = 0.2834678415, which is approximately 0.28. So the answer is 0.28.

Answer:

0.28 (Option D: 0.28)