pg.249 6.1.18 and 19
18.Exactly one of six similar keys opens a certain door. If you try the keys, one after another, what is the expected number of keys that you will have to try before success?
The probability of each key being the right one is \(\frac{1}{6}\), but since the order in which you select them is different, the expected value would be:
\(E(X) = p(x)(x)= \Big(\frac{1}{6}\Big)\Big(1+2+3+4+5+6\Big)=\frac{21}{6}=\frac{7}{2}\)
19. 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. Show that if he just guesses a subset uniformly and randomly his expected score is zero.
So, for this problem, we’re going to first pretend that the correct answer is the first one. Then, because the student can select more than one, we are going to calculate the his score for each possible answer.
{1st}= 3
{2nd}= -1
{3rd}= -1
{4th}= -1
{1st, 2nd}= 3-1 = 2
{1st, 3rd}= 3-1 = 2
{1st, 4th}= 3-1 = 2
{2nd, 3rd}= -1-1 = -2
{2nd, 4th}= -1-1 = -2
{3rd, 4th}= -1-1 = -2
{1st, 2nd, 3rd}= 3-1-1 = 1
{1st, 3rd, 4th}= 3-1-1 = 1
{2nd, 3rd, 4th}= -1-1-1 = -3
{1st, 2nd, 3rd, 4th}= 3-1-1-1 = 0
and, finally, the fifteenth subset is if the student doesn’t choose any answers, which automatically gives them zero points: {none}= 0.
\(E(X) = p(x)(x) = \Big(\frac{1}{15}\Big)\Big(3-1-1-1+2+2+2-2-2-2+1+1+1-3\Big)=\frac{0}{15}=0\)