Question

A problem in mathematics is given to three students whose chances of solving it are respectively \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{4}\). What is the probability that the problem will be solved ?

Answer 1

Let A denotes the problem is solved by first student.
\(\mathrm{B}\) denotes the problem is solved by second student.
\(\mathrm{C}\) denotes the problem is solved by third student.
Clerly, A,B,C are independent events.
We have \(P(A)=\frac{1}{2}\)
\(P(B)=\frac{1}{3}\),and
\(P(C)=\frac{1}{4}\)
Now, \(\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\)
\(=1-\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}})\)
\(=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})\)
\(=1-\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \mathrm{P}(\overline{\mathrm{C}})\) (Since, A,B,C are independent events, so their
compllements also)
\[
\begin{aligned}
&=-1\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \\
&=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}=\frac{3}{4}
\end{aligned}
\]