Question

A problem in statistics is given to three students whose chance of solving it are $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ respectively. The probability that the question is solved is

• A. $\frac{1}{24}$
• B. $\frac{23}{24}$
• C. $\frac{1}{12}$
• D. $1$

Answer 1

The correct option is B $\frac{23}{24}$

Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event

$\therefore A,B,C$ are independent events

Now, $P\left(A^s\right)=$ $\frac{1}{2}$ , P(B) = $\frac{1}{5}$ and P(C)=$\frac{1}{4}$

$P\left(A^s\right)=$$\frac{1}{2}$, $P\left(B^s\right)$=$\frac{2}{3}$, $P\left(C^s\right)=$ $\frac{1}{4}$

$\therefore P\left(nonesolvestheproblem\right)=P\left(notA\right)and\left(notB\right)and\left(notC\right)$

$=P(A\cap B\cap C)$

$=P\left(A\right)P\left(B\right)P\left(C\right)$ [∵ A, B, C are Independent]

=$\frac{1}{2}$$\frac{1}{4}$$\frac{1}{3}$

=$\frac{1}{24}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1-\frac{1}{24}$ = $\frac{23}{24}$