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 ?

• A. $\frac{1}{4}.$
• B. $\frac{3}{4}.$
• C. $\frac{1}{2}.$
• D. $\frac{1}{3}.$

Answer 1

The correct option is B $\frac{3}{4}.$

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