Probability that at least $4$ difficult exams will occur before a fair one An exam is classified as difficult with probability $p$ or fair with probability $q$ where $p+q=1$. Exams are taken one after the other. What is the probability that at least $4$ difficult exams will occur before the first fair one?
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Let $X$ be the number of difficult exams before a fair exam. $X$ is geometric with success probability $q$. We want to find $P(X \geq 4)$ which means
$$ P(X \geq 4) = \sum_{x=4}^{\infty} (1-q)^{x-1} q $$
which equals
$$ (1-p) \sum_{x=4}^{\infty} \frac{1}{p} p^x = \frac{1-p}{p} \sum_{x=0}^{\infty} p^x \cdot p^{4} = p^3(1-p) \cdot \frac{1}{1-p} = p^3 $$
is this correct?
 A: As @lulu noticed, your answer isn't correct.
A geometric distribution $X$ with success probability $q$, like the one in your problem, is defined by
$$ P(X\!=\!k)\,=\,q(1-q)^k \qquad k\ge 0$$
So, if you want at least 4 unsuccessful attempts
$$ P(X\!\ge\!4)\,=\,\sum_{k=4}^\infty P(X\!=\!k) = \sum_{k=4}^\infty q(1-q)^k = \dots = p^4$$
I guess my convention for geometric distribution is slightly different from yours:


*

*$ P(X\!=\!k)\,=\,q(1-q)^k $, the one I used, is the probability of $k$ failures before the first success and its support is $k\ge 0$

*$ P(Y\!=\!k)\,=\,q(1-q)^{k-1} $, the one you used, is the probability of the first success on the $k$-th attempt, where $k\ge 1$


They're the same distribution, but you must pay attention to the index $k$. Using the second convention in your problem means that you're looking for a success on the 5th attempt or later:
$$ P(Y\!\ge\!5)\,=\,\sum_{k=5}^\infty P(Y\!=\!k) = \sum_{k=5}^\infty q(1-q)^{k-1} = \dots = p^4 $$
As you can see, the result is the same.
