student $Y$ solves the same problem is $\frac{1}{4}$. What is the probability that
i. the problem is not solved?
ii. the problem is solved?
iii. the problem is solved exactly by one of them?
$\begin{aligned} & \therefore P(A)=\frac{2}{5}, P(B)=\frac{1}{4} \\ & \therefore P\left(A^{\prime}\right)=1-P(A)=1-\frac{2}{5}=\frac{3}{5} \\ & P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{3}{4}\end{aligned}$
Since A and B are independent events, A’ and B’ are also independent events.
(i) Let event C: Problem is not solved.
$\begin{aligned} & \therefore P(C)=P\left(A^{\prime} \cap B^{\prime}\right) \\ & =P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right) \\ & =\frac{3}{5} \times \frac{3}{4} \\ & =\frac{9}{20}\end{aligned}$
Let event b: Problem is solved.
Problem can be solved if at least one of the two students solves the problem.
∴ P(D) = P(at least one student solves the problem)
= 1 – P(no student solves the problem)
= 1 – P(A’ ∩ B’)
= 1 – P(A’) P(B’)
$\begin{aligned} & =1-\frac{3}{5} \times \frac{3}{4} \\ & =1-\frac{9}{20} \\ & =\frac{11}{20}\end{aligned}$
Let event c: The problem is solved exactly by one of them.
∴ P(E) = P(A’ ∩ B) ∪ P(A ∩ B’)
= P(A’) . P(B) + P(A) . P(B’)
$\begin{aligned} & =\left(\frac{3}{5} \times \frac{1}{4}\right)+\left(\frac{2}{5} \times \frac{3}{4}\right) \\ & =\frac{3}{20}+\frac{6}{20} \\ & =\frac{9}{20}\end{aligned}$
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(i) P(A’/B)