Binomial distribution with 3 different probability

Question

$20$ questions in a test. The probability of getting correct first $10$ questions is $1$. The probability of getting correct next $5$ questions is $\frac 13$. The probability of getting correct last $5$ questions is $\frac 15$. What is the probability of getting exactly $11$ questions correctly?

This is the question. I don't know how to calculate this question.

I tried $$1* {}^5C_1*{\frac 13}\left(\frac 23\right)^4*\left(\frac 45\right)^5+1*\left(\frac 23\right)^5*{}^5C_1*\left(\frac 15\right)\left(\frac 45\right)^4$$

But I am not sure the answer.

What if asking futher about expectation and variance? that will be a mess

Answer 1

For $k<0$:

$P(k)=0.$

For $k\geq10$:

$$P(k)=\sum_{i=\max(k-15,0)}^{\min(k-10,5)}p(i,k-10-i)$$ Where $$p(i,j)={5 \choose i}\left(\frac{1}{3}\right)^i\left(\frac{2}{3}\right)^{5-i} {5 \choose j}\left(\frac{1}{5}\right)^j\left(\frac{4}{5}\right)^{5-j}$$

Finally: for $k\in \{0,1,2,...,20\}$: $$P(k)=\begin{cases}0&, \text{for }k<10\\ \sum_{i=\max(k-15,0)}^{\min(k-10,5)}p(i,k-10-i), \text{for }k\geq 10\end{cases}$$

For $k=11$ we have: $$P(11)=p(0,1)+p(1,0)=5\frac{2+4}{15}\left(\frac{8}{15}\right)^4=\frac{8192}{50625}\approx 0.16$$

Answer 2

The fact that "The probability of getting correct first 10 questions is 1" means that you need to get exactly one of the last 10 questions correct.

You can do that in either of two ways: 1) Get exactly one of the next 5 questions correct and get all of last 5 incorrect. or 2) Get all of the next 5 questions incorrect and get exactly one of the last 5 correct.

1: The probability of getting any of the next 5 questions correct is 1/3, the probability of getting any incorrect is 2/3. The probability of getting exactly one correct and the other 4 incorrect, using the binomial probability formula, is $5(1/3)(2/3)^4= \frac{80}{243}$. The probability of getting any one of the last 5 questions correct is 1/5, the probability of getting a question incorrect is 4/5. The probability of getting all 5 wrong is $(4/5)^5= \frac{1024}{3125}**.

The probability of (1), of getting exactly one of the next 5 questions correct and getting all 5 of the last 5 questions incorrect is $\frac{80}{243}\frac{1024}{3125}$

2) The probability of getting all 5 of the next 5 questions incorrect is $(2/3)^5= \frac{32}{243}$ and the probability of getting exactly one of the last 5 correct is $5(1/5)(4/5)^4= \frac{1024}{3125}$.

The probability of (2), of getting all of the next 5 questions wrong and exactly one of the last 5 correct, is $\frac{32}{243}\frac{1024}{3125}$.

The probability of one of those two happening is the sum of those two probabilities.