Conditional probability - A problem is being solved by 3 students independently. If the problem is solved, what is the probability of there being at least one student that didn't solve the assignment

Question

The title is lacking some details because of the word limit. Here is the full text of the problem:

One problem is being solved by 3 students independently of each other. The probabilities of them solving the problem are 0.84, 0.7 and 0.6. If the problem is solved, what is the probability of there being at least one student that didn't solve the assignment

My (attempt at a) solution to the problem is as follows:

First, the probability of the solution being solved is:

P(X)=P(ABC)+P(~ABC)+P(A~BC)+P(AB~C)+P(~A~BC)+P(~AB~C)+P(A~B~C)=1-P(~A~B~C)=0.9808

Second, the probability of there being at least one student that has not solved the assignment:

P(Y)=P(~ABC)+P(A~BC)+P(AB~C)+P(~A~BC)+P(~AB~C)+P(A~B~C)+P(~A~B~C)=0.16*0.7*0.6+0.84*0.3*0.6+0.84*0.7*0.4+0.16*0.3*0.6+0.16*0.7*0.4+0.84*0.3*0.4+0.16*0.3*0.4=0.6472

Now, using conditional probability formula, I'd calculate the probability of there being at least one student that has not solved the assignment given that the assignments is solved as follows:

P(Y|X)=P(Y∩X)/P(X)=(P(~ABC)+P(A~BC)+P(AB~C)+P(~A~BC)+P(~AB~C)+P(A~B~C))/P(X)=0.628/0.9808=0.640293638

Could someone confirm or refute my solution?

Answer 1

There's a much simpler way:

The probability of all students being able to solve the problem is 0.84×0.7×0.6=0.3528.

Therefore 1-0.3528=0.6472 is the probability that at least one student wasn’t able to solve the problem.

Since you came to the same conclusion, you're correct if I'm correct.

HOWEVER:

The question states that the problem was solved and we know that the probability of no-one solving it is 0.16×0.3×0.4=0.0192. The probability of someone solving it is therefore 1-0.0192=0.9808. The probability of 0.6472 includes the outcome that no-one solved the problem, and this has to be discounted. Doesn't that mean that we need to deduct 0.0192 from 0.6472=0.628?

SUMMARY

Consider all possible outcomes:

(1) All 3 students solved the problem;

(2) Some students solved the problem, some didn't; (in this question, some means one or two)

(3) None of the students solved the problem.

The sum total of these probabilities=1. When we exclude (1) and (3) we get what was asked for, so we just have to calculate probabilities for (1) + (3) and subtract from 1. 1-0.3528-0.0192=0.628.

Comments

Yes, I think that makes sense. I was focusing more on following the formula and neglected to think about the problem logically. Thank you for the correction, I appreciate the help!

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Upon further thought, I think we came to the same solution, but in different ways. I also got 0.628 for P(Y∩X) for when one or two students solved the problem and 0.9808 for P(X) for when at least one student solved the problem. Finally, these two results need to be divided as we are looking for the frequency of one or two students solving the problem -P(Y∩X),  compared to the space of events where at least one student solved the problem - P(X). I hope that made sense.

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Yes, I wasn't sure how the given condition that the problem was solved was to be viewed and taken into account when working out the probability.