First i want to say there are a lot of questions related to this, but i couldn't find a similar case.
Suppose we have the typical problem where we need to compute the probability of pass a multiple-choice test. There are 8 questions with 5 options each, only one is correct. 4 of this questions are easy and the probability of do it well is $2/3$, the other 4 are hard and the probability of do it well is $1/5$. You can pass the test if answer correct at least 5 questions.
Im thinking in the binomial distribution here (from 5 to 8), but in this problem the probability of success is not the same for every question (like in similar problems). So maybe i should use the union of the probability of success for any type of answer (easy or hard), and then apply the binomial distribution.
Best Answer
I think David's approach is correct.
This problem is about the sum of two independent binomial random variables.
Let $X$ be the number of easy questions answered correctly, and $P(X = x)$ the probability of getting x easy answers right. Let $Y$ be the number of hard questions answered correctly, and $P(Y = y)$ the probability of getting $y$ hard answers right.
Then define $Z = X + Y$, and let $P(Z = z)$ be the probability of getting $z$ answers right overall.
Then, under the assumption that $X$ and $Y$ are independent,
$$ P(Z = z) = \Sigma_{x=0}^4 P(X = x) * P(Y = z - x).$$
This is merely describing more formally how I suspect David arrived at his answers for $P(Z = 8)$ and $P(Z = 7)$.
Spelling out the method for $P(Z = 6)$:
$$P(Z = 6) = P(X = 4 \text{ and } Y = 2) + P(X = 3 \text{ and }Y = 3) + P(X = 2 \text{ and } Y = 4) $$
$$= (2/3)^4 6 (1/5)^2 (4/5)^2 + 4 (2/3)^3 (1/3) 4 (1/5)^3 (4/5) + 6 (2/3)^2 (1/3)^2 (1/5)^4. $$
Follow the same method for $P(Z=5)$. The probability of passing the test is:
$$P(pass) = P(Z=5) + P(Z=6) + P(Z=7) + P(Z=8).$$
I won't write down the whole calculation, but with the help of R (https://www.r-project.org), I worked out that the probability of passing this test is $0.196 $ (to $3$ d.p.) .