A bag contains 20 fruits of three types: apples, bananas, and oranges. Each type appears in a different quantity. The probability of picking any type of fruit is greater than \frac{1}{5} but less than \frac{1}{2} . Select for Certain a number that must represent the count of one type of fruit in the bag, and select for Minimum a number that represents the smallest possible count of a type of fruit. Make only two selections, one in each column.

Official Solution: Since the probability of picking any type of fruit is greater than \frac{1}{5} but less than \frac{1}{2} , the count of each type must be more than 4 and less than 10. This means each type has a count of 5, 6, 7, 8, or 9. If the highest count is 9, the next highest can only be 6 (any other value would leave less than 5 for the third type), making the third count 5. If the highest count is 8, the next highest can only be 7, making the third count 5. The highest count cannot be 7 or less because, in that case, the total cannot reach 20 (e.g., 7 + 6 + 5 = 18). Hence, the only possible distributions of fruits are (9, 6, 5) or (8, 7, 5). Thus, a number that must represent the count of one type of fruit in the bag is 5, and the smallest possible count of a type of fruit is also 5. Correct answer: Certain "5" Minimum "5"

1. The question asks us to calculate the exact number of one of the fruits and the smallest number of fruits possible . 2. What we're trying to find is very vague, so, let's start by figuring out the restrictions. 3. Let the number of apples, bananas, and oranges be equal to a, b, and c, respectively. a,b,c are all different, a + b + c = 20, and \frac{1}{5} 20, which doesn't work. So, at least one fruit certainly has a quantity of 5. 5. Our answer will be: Certain - 5 and Minimum - 5 .