Selecting the maximum choice

As a supermarket assistant you have to grade some items in an aisle. You have a maximum of 4 items to grade during your current shift. As a lazy assistant, you wish to finish as quick as possible. Out of these 4 items you are asked to grade 2 separately out of 4 marks and the other two out of 1 mark. Another assistant passes by and makes a guess at which two items will be marked out of 4.

1. Find the maximum total marks the other assistant can guess at out of 2 items.
2. Find the expected value and variance.

For the first question, I know that the maximum marks given to these items are: Max(1,4) = 5, Max(1,1) = 2, Max(4,4) = 8.

So I have to find the probability X = 2, 5, 8.

I thought that we could select them this way i.e. P(X=2) = $2/4 \cdot 1/3$, given that we have 2 out of 4 questions, and 1/3 that their maximum is equal to 2. However, I'm unsure of calculating the other probabilities.

Solution 1:

The probability of each case can be calculated using combinatorics and counting.

For the case you did where they are both graded out of 1, the probability of that occuring is the number of ways to choose 2 that are graded out of 1, divided by the total number of ways to choose 2 items:

$$P(X=2)=\frac{\binom{2}{2}}{\binom{4}{2}}=\frac{1}{6}$$

The probability of getting two that are graded out of 4 is calculated the same way:

$$P(X=8)=\frac{\binom{2}{2}}{\binom{4}{2}}=\frac{1}{6}$$

We can use the same process for the one-of-each case. We choose one of the two graded out of 1, and one of the two graded out of 4:

$$P(X=5)=\frac{\binom{2}{1}\binom{2}{1}}{\binom{4}{2}}=\frac{4}{6}$$

Thus our expected value can be calcluated as

$$E(X)=2\frac{1}{6}+5\frac{4}{6}+8\frac{1}{6}=5$$