At breakfast, lunch, and dinner, Joe randomly chooses with equal probabilities either an apple, an orange, or a banana to eat. On a given day, what is the probability that Joe will eat at least two different kinds of fruit?
The only way for Joe not to eat at least two different kinds of fruit is if he eats all apples, all oranges, or all bananas. The probability that he eats all apples is $\left( \frac{1}{3} \right) ^3=\frac{1}{27}$. The probability that he eats all oranges and the probability that he eats all bananas are the same. So, the probability that Joe eats at least two different kinds of fruit is $1-\frac{1}{27}-\frac{1}{27}-\frac{1}{27}=\boxed{\frac{8}{9}}$.
\frac{8}{9}
Level 4
Counting & Probability
Problem: At breakfast, lunch, and dinner, Joe randomly chooses with equal probabilities either an apple, an orange, or a banana to eat. On a given day, what is the probability that Joe will eat at least two different kinds of fruit?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that Joe eats at least two different kinds of fruit, we first determine the total number of possible outcomes for his fruit choices across three meals. Since Joe has three options (apple, orange, banana) for each meal and he eats three meals, the total number of combinations is calculated as follows:

Total outcomes = 3 (choices for breakfast) × 3 (choices for lunch) × 3 (choices for dinner) = 3^3 = 27.

Next, we need to calculate the number of outcomes where Joe eats only one type of fruit throughout the day. There are three scenarios for this: he could eat only apples, only oranges, or only bananas. Thus, the number of outcomes where he eats only one type of fruit is:

Outcomes with one type of fruit = 1 (all apples) + 1 (all oranges) + 1 (all bananas) = 3.

Now, we can find the number of outcomes where Joe eats at least two different kinds of fruit by subtracting the outcomes with only one type of fruit from the total outcomes:

Outcomes with at least two types of fruit = Total outcomes - Outcomes with one type of fruit = 27 - 3 = 24.

Finally, we calculate the probability that Joe eats at least two different kinds of fruit by dividing the number of favorable outcomes by the total number of outcomes:

Probability = Outcomes with at least two types of fruit / Total outcomes = 24 / 27.

To simplify this fraction, we divide both the numerator and the denominator by 3:

Probability = 8 / 9.

Thus, the final answer is $\frac{8}{9}$.<|END_OF_LATENT|> The final answer is $\frac{8}{9}$.