Question

In a 12 item true false exam a student guesses randonmly on each question. To pass the student needs to get at least 7 correct answers what is the proababilty that the student will pass?

Answer 1

Any help :)

Answer 2

exactly 7 is given by \(\binom{12}{7}\left(\frac{1}{2}\right)^{12}\)

Answer 3

where \(\binom{12}{7}\) read "12 choose 7" is the number of way to pick 7 out of a set of 12
sometimes written as \(_{12}C_7\) and best computed with a calculator

Answer 4

repeat for 8, 9, 10, 11, 12
\[\binom{12}{8}\left(\frac{1}{2}\right)^{12}\] and so on

Answer 5

alternatively, take this number
http://www.wolframalpha.com/input/?i=%2812+choose+7%29%2B%2812+choose+8%29%2B%2812+choose+9%29%2B%2812+choose+10%29%2B%2812+choose+11%29+%2B%2812+choose+12%29
and multiply it by \(\left(\frac{1}{2}\right)^{12}\)

Answer 6

i get this
http://www.wolframalpha.com/input/?i=%28%2812+choose+7%29%2B%2812+choose+8%29%2B%2812+choose+9%29%2B%2812+choose+10%29%2B%2812+choose+11%29+%2B%2812+choose+12%29%29%281%2F2%29^%2812%29

Answer 7

This is a binomial distribution, no?

Answer 8

The way you would look at this problem simply is:\[
\sum_{k=7}^{12}P(X=k) = \sum_{k=7}^{12}{12 \choose k}\left(\frac 12\right)^{k}\left(\frac 12\right)^{12-k} = \sum_{k=7}^{12}{12 \choose k}\left(\frac 12\right)^{12}
\]