According to a theory in​ genetics, if tall and colorful plants are crossed with short and colorless​ plants, four types of plants will​ result: tall and​ colorful, tall and​ colorless, short and​ colorful, and short and​ colorless, with corresponding probabilities of 

nine sixteenths916​, 

three sixteenths316​, 

three sixteenths316​, 

and 

one sixteenth116. 

In a multinomial​ experiment, if 

1010 

plants are​ selected, find the probability that 

33 

will be tall and​ colorful, 

22 

will be tall and​ colorless, 

33 

will be short and​ colorful, and 

22 

will be short and colorless.

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Part 1

The probability is 

enter your response here.

​(Round to three decimal places as​ needed.)

To solve this problem, we'll use the formula for the probability of a specific outcome in a multinomial experiment. The multinomial probability formula is:

\[
P(X_1 = x_1, X_2 = x_2, \dots, X_k = x_k) = \frac{n!}{x_1! \times x_2! \times \dots \times x_k!} \times p_1^{x_1} \times p_2^{x_2} \times \dots \times p_k^{x_k}
\]

Where:
- \( n \) is the total number of trials (plants in this case),
- \( x_1, x_2, \dots, x_k \) are the specific outcomes we're interested in,
- \( p_1, p_2, \dots, p_k \) are the probabilities associated with each outcome.

Given:
- \( n = 10 \) (total plants),
- \( x_1 = 3 \), \( x_2 = 2 \), \( x_3 = 3 \), \( x_4 = 2 \),
- \( p_1 = \frac{9}{16} \), \( p_2 = \frac{3}{16} \), \( p_3 = \frac{3}{16} \), \( p_4 = \frac{1}{16} \).

Let's calculate the probability.

\[
P(3, 2, 3, 2) = \frac{10!}{3! \times 2! \times 3! \times 2!} \times \left(\frac{9}{16}\right)^3 \times \left(\frac{3}{16}\right)^2 \times \left(\frac{3}{16}\right)^3 \times \left(\frac{1}{16}\right)^2
\]

1. **Calculate the factorials:**
   \[
   10! = 3,628,800, \quad 3! = 6, \quad 2! = 2
   \]
   \[
   3! \times 2! \times 3! \times 2! = 6 \times 2 \times 6 \times 2 = 144
   \]

2. **Calculate the combination coefficient:**
   \[
   \frac{10!}{144} = 25,200
   \]

3. **Calculate the probabilities raised to the respective powers:**
   \[
   \left(\frac{9}{16}\right)^3 = \frac{729}{4096}, \quad \left(\frac{3}{16}\right)^2 = \frac{9}{256}, \quad \left(\frac{3}{16}\right)^3 = \frac{27}{4096}, \quad \left(\frac{1}{16}\right)^2 = \frac{1}{256}
   \]

4. **Multiply all the probabilities together:**
   \[
   \frac{729}{4096} \times \frac{9}{256} \times \frac{27}{4096} \times \frac{1}{256} = \frac{177147}{2^{36}}
   \]
   This simplifies to:
   \[
   \frac{177147}{2^{36}} \approx 6.7108864 \times 10^{-6}
   \]

5. **Final probability:**
   \[
   P(3, 2, 3, 2) = 25,200 \times 6.7108864 \times 10^{-6} \approx 0.169
   \]

Thus, the probability is approximately \( \boxed{0.169} \) (rounded to three decimal places).

Would you like more details or have any questions? Here are some related questions:

1. How would the probability change if 15 plants were selected instead of 10?
2. What is the expected number of tall and colorful plants out of 10?
3. How would you calculate the probability if the selection numbers were 4, 3, 2, 1?
4. How does the multinomial distribution relate to the binomial distribution?
5. What are some real-world applications of the multinomial distribution?

**Tip:** In multinomial experiments, the order of outcomes within the same category does not matter, so the formula accounts for different combinations by dividing by factorials.

Learn how to calculate probabilities in a multinomial experiment using the example of genetic cross outcomes. This problem involves determining the probability of specific plant types based on genetic traits and their respective probabilities. Useful for understanding multinomial distributions and applying probability theory.

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