Question

A problem in mathematics is given to $4$ students whose chances of solving individually are $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac{1}{5}$. The probability that the problem will be solved at least by one student is

• A. $\frac{2}{3}$
• B. $\frac{3}{5}$
• C. $\frac{4}{5}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{4}{5}$

Let $A,B,C,D$ be the events that $1^{st},2^{nd},3^{rd},4^{th}$ student solves the problem.

Probability that no student solves the problem is
$P\left(\overline{A}\right)\cdot P\left(\overline{B}\right)\cdot P\left(\overline{C}\right)\cdot P\left(\overline{D}\right)=$
$[1-\frac{1}{2}][1-\frac{1}{3}][1-\frac{1}{4}][1-\frac{1}{5}]$
$=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}=\frac{1}{5}$

$\Rightarrow$ Probability that the problem will be solved by at least one student is equal to $1-\frac{1}{5}=\frac{4}{5}$