A multiple choice exam is given. A problem has four possible answers, and exactly one answer is correct. The student is allowed to choose a subset of the four possible answers as his answer. If his chosen subset contains the correct answer, the student receives three points, but he loses one point for each wrong answer in his chosen subset. Determine the expected score.

• 0
• 0.5
• 1
• 1.5
• 2

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: A

The number of subsets of size [math]k[/math] containing the answer equals [math]\binom{3}{k-1}[/math] and the points associated with each such subset is [math]4-k[/math]. The number of subsets of size [math]k[/math] not containing the answer equals [math]\binom{3}{k}[/math] and the points associated with each such subset equals [math]-k[/math]. Hence the expected value equals

[[math]] \sum_{k=1}^4 (4-k) \binom{3}{k-1} - \sum_{k=1}^4 k\binom{3}{k} [[/math]]

Now we use the binomial identity:

[[math]] \binom{4}{k} = \binom{3}{k-1} + \binom{3}{k} [[/math]]

and obtain an expected value of

[[math]] \sum_{k=1}^4 4\binom{3}{k-1} - \sum_{k=1}^4 k\binom{4}{k}. [[/math]]

But we also have

[[math]] 4 \binom{3}{k-1} = \frac{ 4!}{(4-k)!)(k-1)!} = k\binom{4}{k} [[/math]]

which implies that expected value equals zero as desired.