A problem in Mathematics is given to three students whose chances of solving it are $\dfrac{1}{2}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$.
- What is the probability that the problem is solved?
- What is the probability that exactly one of them will solve it?
Let $A$ be the event that the problem is solved by the first student.
Let $B$ be the event that the problem is solved by the second student.
Let $C$ be the event that the problem is solved by the third student.
Given: $\;$ $P \left(A\right) = \dfrac{1}{2}$, $\;$ $P \left(B\right) = \dfrac{1}{3}$, $\;$ $P \left(C\right) = \dfrac{1}{4}$
Then, $P \left(\overline{A}\right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$, $\;$ $P \left(\overline{B}\right) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$, $\;$ $P \left(\overline{C}\right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$
- Probability that the problem is solved:
$\begin{aligned} P \left(\text{problem is solved}\right) & = P \left(\text{problem is solved by at least one of them}\right) \\\\ & = P \left(A \cup B \cup C\right) \\\\ & = 1 - P \left(\overline{A \cup B \cup C}\right) \\\\ & = 1 - P \left(\overline{A} \cap \overline{B} \cap \overline{C}\right) \;\;\; \left[\text{By DeMorgan's law}\right] \\\\ & = 1 - P \left(\overline{A}\right) \times P \left(\overline{B}\right) \times P \left(\overline{C}\right) \\\\ & \left[\because \; \text{A, B, C are independent events, } \right. \\ & \hspace{2cm} \left. \overline{A}, \overline{B}, \overline{C} \text{ are also independent events}\right] \\\\ & = 1 - \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \\\\ & = \dfrac{3}{4} \end{aligned}$
- Exactly one of them solves the problem:
$P \left(\text{exctly one solves the problem}\right)$
$= P \left[\left(A \cap \overline{B} \cap \overline{C}\right) \cup \left(\overline{A} \cap B \cap \overline{C} \right) \cup\left(\overline{A} \cap \overline{B} \cap C\right)\right] $
$ = P \left(A \cap \overline{B} \cap \overline{C}\right) + P \left(\overline{A} \cap B \cap \overline{C}\right) + P \left(\overline{A} \cap \overline{B} \cap C\right) $
$ \left[\because \; \left(A \cap \overline{B} \cap \overline{C}\right), \; \left(\overline{A} \cap B \cap \overline{C}\right), \; \left(\overline{A} \cap \overline{B} \cap C\right) \right.$
$\hspace{3cm} \left. \text{ are mutually exclusive events} \right]$
$= P \left(A\right) \times P \left(\overline{B}\right) \times P \left(\overline{C}\right) + P \left(\overline{A}\right) \times P \left(B\right) \times P \left(\overline{C}\right) $
$\hspace{5cm} + P \left(\overline{A}\right) \times P \left(\overline{B}\right) \times P \left(C\right) $
$ \left[\because \; \text{A, B, C are independent events, } \overline{A}, \overline{B}, \overline{C} \text{ are also independent events}\right] $
$ = \left(\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4}\right) + \left(\dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{3}{4}\right) + \left(\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{1}{4}\right) $
$ = \dfrac{11}{24} $