Question

In a multiple-choice test, each question has four options. Students will get 10 points for each correct answer; lose four points for each incorrect answer; and receive no points for unanswered questions. A student does not know the correct answer for one question. Is it to her advantage or disadvantage to guess an answer? Show your calculations for expected value.

Answer 1

Think of correct answers as worth 10, incorrect answers worth -4 and blanks worth zero. The (probability of being correct times the value of being correct) plus the (probability of being incorrect times the value of being incorrect) needs to be better than zero for guessing to work.

Answer 2

another way to do this is imagine you answer all four, and see what score you get
you get 10 points for the correct answer, but lose 12 points for the three incorrect answers, for a net loss of 2 points,
so it is not good to guess

Answer 3

of course that does not show your calculation for expected value
to calculate that one, you need the probability you guess correctly, which is \(\frac{1}{4}\) and also the probability that you guess incorrectly, which is evidently therefore \(\frac{3}{4}\)
now lets imagine you guess.
one fourth of the time you get it right, for a total of ten points, but three fourths of the time you guess wrong and lose 4 points
the expected value is therefore
\[10\times \frac{1}{4}-4\times \frac{3}{4}\]

Answer 4

I did it like this







Well if I look at it as precentages of chance, then it is a disadvantage to guess.

not guessing gives 0 points

If I guess, then there is a 25% chance I get 10 points, and a 75 percent chance I get -4
So for 4 guesses I will get 1 right, and 3 wrong.
The one right counts for 10 points, but the 3 wrong count for -12. (3 x -4 =-12)
Add them up and I would end up with -2 points, so it would have been better to not guess at all.

Answer 5

yes that looks good, although you were asked for the "expected value" so average your loss of -2 points over the 4 answers and you get \(\frac{-2}{4}=-\frac{1}{2}\) meaning you "expect" to lose one half point per guess

Answer 6

the actual "expected value" using percents is \(10\times .25-4\times .75=-.5\)

Answer 7

thank you

Answer 8

yw