Probability of selecting correct answer in 15 out of 25 exercises with 0.25 chance There are 25 exercises, each one consists of answers: a, b, c, d and only one answer is correct. My question is what is the probability of selecting correct answer in 15 out of 25 exercises.
My idea: $\dfrac{25!}{15! * 10!} * \left(\dfrac14\right)^{15}$
$\dfrac14$ is chance of hitting correct answer and multiplied by 15 (number of needed answers) and $\dfrac{25!}{15! * 10!}$ is number of draws of answers.
But the result is a very small number (chance is almost zero which in my opinion is logically incorrect), so I think theres a error in my thinking. Please tell me what did I do wrong. 
 A: The correct answer is actually even smaller than yours: it’s
$$\binom{25}{15}\left(\frac14\right)^{15}\left(\frac34\right)^{10}\approx0.00017\;.$$
The factor of $\binom{25}{15}$ is the number of ways of choosing which $15$ of the $25$ answers are correct; $\left(\frac14\right)^{15}$ is the probability that these $15$ answers are all correct; and $\left(\frac34\right)^{10}$ is the probability that all $10$ of the other answer are incorrect.
A: Your formula is missing a portion.
Suppose, for a moment, we calculate the probability you get the first 15 right, and the rest wrong. Not only do you need the $\frac{1}{4}^{15}$ probability for each of the correct answer, but you need the $\frac{3}{4}^{10}$ probability for the wrong answers.
So the probability for that single event is $(\frac{1}{4})^{15}\cdot (\frac{3}{4})^{10}$
But as you recognized, these 15 correct answers could be distributed among the 25 in any manner, so you need to count the ${25 \choose 15}$ ways that this can happen.
The net result is then ${25 \choose 15}(\frac{1}{4})^{15}\cdot (\frac{3}{4})^{10}$.
That it is so small is not too surprising: after all, this is a binomial distribution, so its expected value is $25*(1/4)\approx 6$ with a standard deviation $\sqrt{25*0.25*0.75}\approx 2$, and we will be expecting most (at least 75%, but much more in this case) of the probability density to be between 2 and 10. In fact, a little more than 96% is between 2 and 10.
This only leaves less than 4% to be split below 2 and above 10. Compuations show 0.7% lies below 2, and 2.9% lies above.
