A student taking a MC exam with 6 QQ's in which each equation has 4 options, and the student selects the options randomly.
What is the probability that she will get 6 qq's correct? Round to 4 decimal places as needed.
Can someone please explain to me, on how to work this out? I'm quite confused ~~".
0.25% chance in getting one qq right.
0.75% in getting one qq wrong.
This is an example of the binomial distribution.
Binomial because, for each question, there are two possible outcomes, R-Right and W-Wrong.
P(R) = 0.25
P(W) = 0.75
For 6 right answers it cannot be P(one right) times 6 (= 1.5) because you cannot have a probabliity over 1.
If adding each probability were correct it would get more and more likely to get none wrong, the more questions that were asked. ????
No, what you do is multiply the probabilities like this:
P(RR) = 0.25 x 0.25
P(RRR) = 0.25 x 0.25 x 0.25
and so on.
So for 6 questions all correct
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei