Calculate the probability of solving test positive

Question

A test contains 10 questions, each with 5 possible answers but only one of them is correct. Test is positive if at least half of the questions are correct answered.   Calculate the probability of solving test positive if person just randomly answered the questions.

Solution: $\frac{320249}{9765625}$

I was tyring to use binomial distribution with $n=10, k=5$, and $p=0.2$ but I can't get correct solution. Where is my mistake?

Answer 1

Let $X$ be the number of correctly answered questions, then $ X \sim Bin(10, 1/5)$, hence the probability of a "positive test" can be calculated by $$ P(X\ge 5) = \sum_{k=5}^{10}\binom{10}{k}\frac{1}{5^k}\times\frac{4^{10-k}}{5^{10 - k}} $$

Answer 2

If you want it a little bit more developed this may help you. First, all possible ways the exam can be answered: each question has 5 different answers and there are in total 10 questions:

$$5^{10}= 9,765,625$$

Then, all the possibilities of a positive test

• 5 questions with the right answer and 5 other with a wrong answer. One case is: $$1\cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4$$ And since the order matters (is not the same to miss question 3 than to miss question 7) we have to consider all possible permutations of different elements:

$$4^5 \dfrac{10!}{5!5!}$$

• 6 questions with the right answer and 4 other with a wrong answer:

$$4^4 \dfrac{10!}{6!4!}$$

...

• 10 questions right

$$1$$

Adding all these possible positive tests: $258,048+53,760+7,680+720+40+1 = 320,249$. And probability:

$$P = \dfrac{\text{positive outcomes}}{\text{all possible outcomes}} = \dfrac{320,249}{9,765,625}$$

Answer 3

I think you are taking only case with 5 correct answers. But here you have to take cases with 5, 6, 7, 8, 9 and 10 correct answers.

Then easy method to do is find probability with cases 0, 1, 2, 3 and 4 correct answers. And subtract this probability from 1. You get final answer.