Question

A problem in Mathematics is given to 4 students. A, B, C and D. Their chances of solving the problems respectively are $\frac{1}{3},\frac{1}{4},\frac{1}{5}$ and $\frac{2}{3}$. What is the probability that (i) the problem will be solved ? (ii) at most one of them will solve the problem ?

Answer 1

Let A, B, C and D denote the events that A, B, C and D solves the problem in mathematics respectively.
$\therefore P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{4},P\left(C\right)=\frac{1}{5}andP\left(D\right)=\frac{2}{3}\Rightarrow P\left(\overline{A}\right)=\frac{2}{3},P\left(\overline{B}\right)=\frac{3}{4},P\left(\overline{C}\right)=\frac{4}{5},\text{and}P\left(\overline{D}\right)=\frac{1}{3}.$
(i) P(the problem will be solved) = $P\left(A\right)=P(A\cup B\cup C\cup D)=1-P(\overline{A}\cap \overline{B}\cap \overline{C}\cap \overline{D})\Rightarrow =1-P\left(\overline{A}\right)\times P\left(\overline{B}\right)\times P\left(\overline{C}\right)\times P\left(\overline{D}\right)=1-\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{1}{3}=\frac{13}{15}.$
(ii) P(at most one of them will solve the problem)
$=P(\overline{A}\cap \overline{B}\cap \overline{C}\cap \overline{D})+P(A\cap \overline{B}\cap \overline{C}\cap \overline{D})+P(\overline{A}\cap B\cap \overline{C}\cap \overline{D})+P(\overline{A}\cap \overline{B}\cap C\cap \overline{D})+P(\overline{A}\cap \overline{B}\cap \overline{C}\cap D)\Rightarrow =\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{1}{3}+\frac{1}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{1}{3}+\frac{2}{3}\times \frac{1}{4}\times \frac{4}{5}\times \frac{1}{3}+\frac{2}{3}\times \frac{3}{4}\times \frac{1}{5}\times \frac{1}{3}\times +\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{2}{3}=\frac{49}{90}$