If the probability that A solves a problem is 1/2and that for B is 3/4 and if they aim at solving

Answer 1

I think you are asking one of these two questions.

1. What if they together try to solve the problem? Then what is the probability that they both solve the problem?
2. For this, think it what fraction (probability is a fraction of a scenario to the all possible cases) of a question can be solved by a student, and then for those that are solved by the first student, what fraction can be solved by the other student too. Thus they will both solve it (1/2) * (3/4) = 3/8 of the times. So the probability would be 3/8.
3. What if they together try to solve the problem? Then what is the probability that at least one of them will solve the problem (that means somehow the problem is solved, does not matter who solves it or whether both solve it)?
4. For this, think that there can be 4 scenarios against one question, (a) only A solves it, (b) only B solves it, (c) both solve it, and (d) none solve it. We are interested in knowing the fraction of times (a) or (b) or (c) happens but (d) does not happen. That means the probability would be 1 - the probability of scenario (d) i.e. none solve it. The probability that A does not solve is (1- 1/2), and then for each of those that A does not solve, the probability that B also does not solve is (1 - 3/4). Therefore the probability of none solve it is = (1 - 1/2) * (1 - 3/4) = (1/2) * (1/4) = 1/8. This the desired probability against the question is 1 - scenario (d) probability = 1 - (1/8) = 7/8.

I hope this answers your question. All the best!