Question

There are 10 multiple choice questions:
the first has two options A, B. The second question has three options A, B, C. So on, until the tenth question has eleven options A, B, C, D, E, F, G, H, I, J, K.

If you choose C for all the questions, what is the probable (expected) score (to one decimal place) for the whole set of questions, out of ten?

Answer 1

`the first has two options A, B.`

how can i choose C ?

Answer 2

you'll get the first question wrong,

Answer 3

okay that makes sense :)

Answer 4

p = {0, 1/3, 1/4, 1/4,... 1/11}

Answer 5

so the sum is . . .

Answer 6

can i simply add like that for expected value ? i don't remember, need to review once!

Answer 7

yes
the expected mark will be 0 + 1/3 + 1/4 + 1/5 +...+ 1/11

Answer 8

= 7/12 + 11/30 + ... + 1/11

Answer 9

ahh yes yes ! \(\large \sum \limits_{n=1}^{9}\dfrac{1}{n+2} \approx 1.5\)

Answer 10

http://www.wolframalpha.com/input/?i=+1%2F3+%2B+1%2F4+%2B+1%2F5+%2B...%2B+1%2F11

Answer 11

\[\color{red}\checkmark\]

Answer 12

Is there some technique to solve that sigma sum?

Answer 13

his expected score increases by 33.33% had he selected options 1 or 2

Answer 14

i don't think a closed form exists for a harmonic series
http://mathworld.wolfram.com/HarmonicSeries.html

Answer 15

i guess this series is a difference of harmonic series-es

Answer 16

http://mathworld.wolfram.com/HarmonicNumber.html