Probability that all problems are solved

Question

The problem is

A solves a problem with probability 0.6. B solves a problem with probability 0.5. They are given 7 problems and each chooses 5 out of the 7 problems randomly and solves them independently. What is the probability that all the 7 problems are solved by at least one of them?

It seems like a binomial distribution problem but I am not sure how to do.

I am thinking that probability that a problem is solved

P(solved)=P(A solved)+P(B solved)-P(A and B both solved)
         =(5/7)(0.6)+(5/7)(0.5)-(5/7)(5/7)(0.6)(0.5)
         =31/49

then all problems are solved is

P(all solved)=(31/49)^7

I don't think I am doing right but am not sure where. Can anyone give a thought?

Answer 1

The way I interpret this, if $A$ and $B$ both decided to work on the first five problems... then regardless of their success at solving those first five problems correctly neither of them attempted problems $6$ or $7$ and so neither of those problems are able to be solved. Taking it a step further, we recognize that for all seven problems to have been solved, we require as a first step that $A$ and $B$ collectively attempt all seven of the problems... that is, there are $2$ problems attempted only by $A$, $2$ problems attempted only by $B$, and $3$ problems attempted by both.

Your attempted answer incorrectly used this "each chooses 5 problems to work on" aspect of the problem. Whether or not one question is selected is very much dependent on whether another is selected. Of those who were selected, the problem being successfully completed will be independent however.

Next, I interpret this as a "problem is solved" iff at least one of the students correctly solved it.

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Step 1: Recognize it is possible to have collectively solved all seven problems only if all seven were attempted. Find the probability that the problems attempted collectively cover all seven available problems.

Without loss of generality, suppose $A$ picked what problems to work on first and they happened to be the first five problems. $B$ now picks the problems they work on and we need two of those problems to have been the last two problems. There are $\binom{7}{5}$ ways that $B$ could have picked their problems, only $\binom{5}{3}$ of which result in the last two being picked.

Step 2: For each problem, find the probability it was successfully completed by at least one of the students. Recognize the difference between the questions attempted only by $A$, the questions attempted only by $B$, and the questions attempted by both.

If a question was attempted only by $A$ there is a $0.6$ chance it was completed correctly. If a question was attempted only by $B$ there is a $0.5$ chance it was completed correctly. If a question was attempted by both there is an $0.5 + 0.6 - 0.5\cdot 0.6 = 1.1-0.3=0.8$ chance it was completed correctly by at least one of them.

We get as a final answer then:

$$\dfrac{\binom{5}{3}}{\binom{7}{5}}(0.6)^2(0.5)^2(0.8)^3$$

Answer 2

Considerations - $A$ and $B$ each choose problems independently and work on them independently. Also assuming the answers are objective and they are either right or wrong, nothing in between. The given probability of solving is probability of getting the right answer and if anyone of them get the answer right, that question gets considered solved.

First number of ways to ensure all $7$ questions are chosen between the two. This can only happen if each of them have $3$ questions that is common and $2$ questions that are different from the other person.

So $C(7) = {7 \choose 5} {5 \choose 3} {2 \choose 2}$

$P(7) = \displaystyle \frac{{7 \choose 5} {5 \choose 3} {2 \choose 2}}{{7 \choose 5} {7 \choose 5}} = \frac{10}{21}$

Now probability of $A$ solving a problem $= \frac{3}{5}$

Probability of $B$ solving a problem $= \frac{1}{2}$

Probability that a common question is not solved by either of them $= \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$

So they solve a common question with probability $ = \frac{4}{5}$

So probability of all questions getting solved $= \frac{10}{21} \times (\frac{3}{5})^2 \times (\frac{1}{2})^2 \times (\frac{4}{5})^3$