Question

The probability that a student X solves a problem in dynamics is ${\displaystyle \frac{2}{5}}$ and the probability that student Y solves the same problem is ${\displaystyle \frac{1}{4}}$. What is the probability that

1. the problem is not solved
2. the problem is solved
3. the problem is solved exactly by one of them

Answer 1

Let event A: Student X solves the problem in dynamics,

event B: Student Y solves the problem in dynamics.

∴ P(A) = ${\displaystyle \frac{2}{5}}$, P(B) = ${\displaystyle \frac{1}{4}}$

∴ P(A') = 1 – P(A) = ${\displaystyle 1-\frac{2}{5}=\frac{3}{5}}$

P(B') = 1 – P(B) = ${\displaystyle 1-\frac{1}{4}=\frac{3}{4}}$

A and B are independent events,

A' and B' are also independent events

(i) Let event C: Problem is not solved.

∴ P(C) = P(A' ∩ B')

= P(A') · P(B')

= ${\displaystyle \frac{3}{5}\times \frac{3}{4}}$

= ${\displaystyle \frac{9}{20}}$

(ii) Let event D: Problem is solved.

Problem can be solved if at least one of the two students solves the problem.

∴ P(C) = P(at least one student solves the problem)

= 1 - P (no student solves the problem)

= 1 - P(A' ∩ B')

= 1 - P (A') · P (B')

${\displaystyle =1-\frac{3}{5}\times \frac{3}{4}}$

${\displaystyle =1-\frac{9}{20}}$

${\displaystyle =\frac{11}{20}}$

(iii) Let event E: 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')

${\displaystyle =(\frac{3}{5}\times \frac{1}{4})+(\frac{2}{5}\times \frac{3}{4})}$

${\displaystyle =\frac{3}{20}+\frac{6}{20}}$

${\displaystyle =\frac{9}{20}}$