Chef Fabio went to the store and bought 2 cans of beans, 2 cans of tomatoes, and 2 cans of tuna, all of which were the same shape and size. On his way home, it rained hard and the labels became soaked and fell off the cans. He wants to make a dish using one can of each type, and begins to open the cans one at a time. The expected number of cans he will have to open to get one can of each type can be expressed as ab, where a and b are coprime positive integers. What is the value of a+b?

Question

Chef Fabio went to the store and bought 2 cans of beans, 2 cans of tomatoes, and 2 cans of tuna, all of which were the same shape and size. On his way home, it rained hard and the labels became soaked and fell off the cans. He wants to make a dish using one can of each type, and begins to open the cans one at a time. The expected number of cans he will have to open to get one can of each type can be expressed as ab, where a and b are coprime positive integers. What is the value of a+b?

Answer 1

17/7 hence a+b=24

Answer 2

Explain, please?

Answer 3

24 is correct but the fraction given is wrong.

Answer 4

24 is correct but the fraction given is wrong. Also this is a brilliant problem, so stop cheating.

Answer 5

To solve this problem, we can use the concept of the coupon collector's problem. In this problem, imagine that there are n different types of coupons, and you keep collecting coupons at random. The question is: on average, how many coupons do you need to collect before you have at least one of each type?

In our case, the types of coupons correspond to the cans of beans, tomatoes, and tuna. We want to find the average number of cans Chef Fabio needs to open before he gets one can of each type.

Let's analyze the scenario step by step:

1. Opening the first can:
- No matter which can Fabio opens first, it will be a new type, so the current count of types is 1.

2. Opening the second can:
- There are now two possibilities:
a) Fabio opens a can of a new type, so the current count of types is 2.
b) Fabio opens a can of the same type he opened first, so the current count of types remains 1.
- The probability of case (a) is 2/3 (since 2 out of 3 cans are different types).
- The probability of case (b) is 1/3 (since 1 out of 3 cans is the same type).

3. Opening the third can:
- There are three possibilities:
a) Fabio opens a can of a new type, so the current count of types is 3.
b) Fabio opens a can of the same type he opened first, so the current count of types remains 1.
c) Fabio opens a can of the same type he opened second but different from the first one, so the current count of types is 2.
- The probability of case (a) is 3/4 (since 3 out of 4 cans are different types).
- The probability of case (b) is 1/4 (since 1 out of 4 cans is the same type as the first one).
- The probability of case (c) is 1/4 (since 1 out of 4 cans is the same type as the second one).

Now, we can calculate the expected number of cans Fabio needs to open to get one can of each type by summing the expected values of each case:

Expected number = (1 * (1 + 0)) * (2/3) + (2 * (1 + 0 + 1)) * (1/3) * (3/4) = 7/4

Therefore, a = 7 and b = 4. The sum a + b = 7 + 4 = 11.

Hence, the value of a + b is 11.