Answer :
- Calculate marginal probabilities for fur color: $P(B_{Fur}) = P(W_{Fur}) = \frac{1}{3}$.
- Calculate marginal probabilities for eye color: $P(B_{Eyes}) = P(R_{Eyes}) = \frac{1}{3}$.
- Check if $P(B_{Fur}, B_{Eyes}) = P(B_{Fur}) * P(B_{Eyes})$: $\frac{1}{6} \neq \frac{1}{9}$.
- Fur color and eye color are not independent events: $\boxed{False}$.
### Explanation
1. Understand the problem
We are given the predicted fractions for different combinations of fur and eye colors. We need to determine if the fur color and eye color are independent events.
2. Plan the solution
To determine independence, we need to calculate the marginal probabilities for fur color and eye color.
3. Calculate marginal probabilities for fur color
The marginal probability for black fur is the sum of the probabilities of black fur with black eyes and black fur with red eyes: $P(B_{Fur}) = P(B_{Fur}, B_{Eyes}) + P(B_{Fur}, R_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$. Similarly, the marginal probability for white fur is: $P(W_{Fur}) = P(W_{Fur}, B_{Eyes}) + P(W_{Fur}, R_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
4. Calculate marginal probabilities for eye color
The marginal probability for black eyes is the sum of the probabilities of black eyes with black fur and black eyes with white fur: $P(B_{Eyes}) = P(B_{Fur}, B_{Eyes}) + P(W_{Fur}, B_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$. Similarly, the marginal probability for red eyes is: $P(R_{Eyes}) = P(B_{Fur}, R_{Eyes}) + P(W_{Fur}, R_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
5. Check for independence
Now we check for independence. If fur color and eye color are independent, then $P(B_{Fur}, B_{Eyes}) = P(B_{Fur}) * P(B_{Eyes})$. We have $P(B_{Fur}, B_{Eyes}) = \frac{1}{6}$ and $P(B_{Fur}) * P(B_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. Since $\frac{1}{6} \neq \frac{1}{9}$, the events are not independent. We can check the other combinations as well. $P(B_{Fur}, R_{Eyes}) = \frac{1}{6}$ and $P(B_{Fur}) * P(R_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. $P(W_{Fur}, B_{Eyes}) = \frac{1}{6}$ and $P(W_{Fur}) * P(B_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. $P(W_{Fur}, R_{Eyes}) = \frac{1}{6}$ and $P(W_{Fur}) * P(R_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. In all cases, the joint probability is not equal to the product of the marginal probabilities.
6. Conclusion
Since the joint probabilities are not equal to the product of the marginal probabilities, fur color and eye color are not independent events.
### Examples
Understanding independence in events is crucial in genetics. For example, if certain traits are inherited independently, the probability of observing a combination of traits can be calculated by multiplying the individual probabilities. If traits are dependent, the calculation becomes more complex, requiring consideration of conditional probabilities.
- Calculate marginal probabilities for eye color: $P(B_{Eyes}) = P(R_{Eyes}) = \frac{1}{3}$.
- Check if $P(B_{Fur}, B_{Eyes}) = P(B_{Fur}) * P(B_{Eyes})$: $\frac{1}{6} \neq \frac{1}{9}$.
- Fur color and eye color are not independent events: $\boxed{False}$.
### Explanation
1. Understand the problem
We are given the predicted fractions for different combinations of fur and eye colors. We need to determine if the fur color and eye color are independent events.
2. Plan the solution
To determine independence, we need to calculate the marginal probabilities for fur color and eye color.
3. Calculate marginal probabilities for fur color
The marginal probability for black fur is the sum of the probabilities of black fur with black eyes and black fur with red eyes: $P(B_{Fur}) = P(B_{Fur}, B_{Eyes}) + P(B_{Fur}, R_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$. Similarly, the marginal probability for white fur is: $P(W_{Fur}) = P(W_{Fur}, B_{Eyes}) + P(W_{Fur}, R_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
4. Calculate marginal probabilities for eye color
The marginal probability for black eyes is the sum of the probabilities of black eyes with black fur and black eyes with white fur: $P(B_{Eyes}) = P(B_{Fur}, B_{Eyes}) + P(W_{Fur}, B_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$. Similarly, the marginal probability for red eyes is: $P(R_{Eyes}) = P(B_{Fur}, R_{Eyes}) + P(W_{Fur}, R_{Eyes}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
5. Check for independence
Now we check for independence. If fur color and eye color are independent, then $P(B_{Fur}, B_{Eyes}) = P(B_{Fur}) * P(B_{Eyes})$. We have $P(B_{Fur}, B_{Eyes}) = \frac{1}{6}$ and $P(B_{Fur}) * P(B_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. Since $\frac{1}{6} \neq \frac{1}{9}$, the events are not independent. We can check the other combinations as well. $P(B_{Fur}, R_{Eyes}) = \frac{1}{6}$ and $P(B_{Fur}) * P(R_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. $P(W_{Fur}, B_{Eyes}) = \frac{1}{6}$ and $P(W_{Fur}) * P(B_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. $P(W_{Fur}, R_{Eyes}) = \frac{1}{6}$ and $P(W_{Fur}) * P(R_{Eyes}) = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}$. In all cases, the joint probability is not equal to the product of the marginal probabilities.
6. Conclusion
Since the joint probabilities are not equal to the product of the marginal probabilities, fur color and eye color are not independent events.
### Examples
Understanding independence in events is crucial in genetics. For example, if certain traits are inherited independently, the probability of observing a combination of traits can be calculated by multiplying the individual probabilities. If traits are dependent, the calculation becomes more complex, requiring consideration of conditional probabilities.
