Question

Given 100 multiple choice questions,
what is the probability of getting 50 questions right,
if the selection A, B, C, or D is random.

Answer 1

\[\binom{n}{k}p^kq^{n-k}\]

Answer 2

binomial distribution

Answer 3

Multiple*

Answer 4

That equation provided by @amistre64 should pretty much help you solve it.

Answer 5

\[\binom{100}{50}(\tfrac14)^{100}q^{100-50}\] What is q?

Answer 6

q is the complement of p ...

Answer 7

if we have 1/4 chance of success, whats the alternative?

Answer 8

(p+q)^n; such that p+q = 1 for a valid distribution. therefore q = 1-p

Answer 9

3/4

Answer 10

we can expand that to (p+q+r+...+z)^n such that p+q+r+...+z=1 and we can determine a lot more variables :)

yes, q = 3/4

Answer 11

\[\binom{100}{50}(\tfrac14)^{100}(\tfrac14-1)^{100-50}\\~\\
=^{100}C_{50}\times\tfrac1{4^{100}}\times(1-\tfrac14)^{100-50}\\
=\frac{100!}{50!(100-50)!}\times4^{-100}\times(\tfrac34)^{50}\\
=\frac{100!}{(50!)^2}\times4^{-150}\times3^{50}\\
\approx 3.6\times10^{-38}\]

Answer 12

~ so it is improbable ,

thank you

Answer 13

is this for getting exactly 50 or at least 50?

Answer 14

exactly 50

Answer 15

*\[\binom{100}{50}(\tfrac14)^{50}(\tfrac14-1)^{100-50}\\~\qquad\qquad\quad\vdots\\
=\frac{100!}{(50!)^2}\times4^{-100}\times3^{50}\\\]

Answer 16

atleast 50 is also very improbable

Answer 17

atleast 25 gives almost half the area :

Answer 18

i see,

but how do i modify the equation

\[\sum_{k=50}^{100}\binom{100}{k}(\tfrac14)^{k}(\tfrac14-1)^{k-50}\]

?

Answer 19

more than half *

Answer 20

yeah i would have expect to get 25 right

Answer 21

we may play with combinatorics, but not sure if we can simplify it....
we use normal approximation when n is large (n > 30)

Answer 22

we convert the binomial distribution into normal distribution and use the zscore tables to extract the areas

Answer 23

For our present problem, normal approximation would be :

mean \(\mu = np = 100*1/4 = 25 \)
standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{100*1/4*3/4} = 5\sqrt{3}/2\)

Answer 24

Zscore for observation 50 = \(\large \dfrac{50 -26 }{5\sqrt{3}/2} = 5.54 \)

Answer 25

P(X>= 50) :

http://www.wolframalpha.com/input/?i=P%28X+%3E%3D++5.54%29

this is the normal approximation forearlier summation formula : \(\sum_{k=50}^{100}\binom{100}{k}(\tfrac14)^{k}(\tfrac14-1)^{k-50}\)

Answer 26

It really matches quite closely as n is large (100)

Answer 27

the larger the value of n, the lesser is the error...