Question

Assertion :The probability of solving a new problem by 3 students are $\frac{1}{2},\frac{1}{3}$ and $\frac{1}{4}$ respectively. The probability that the problem will be solved by them is $\frac{1}{24}$. Reason: If $A$, $B$ and $C$ are three independent events then the probability of at least one of them happening = $1-P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(\overline{C}\right)$

• A. Assertion is True, Statement-2 is True; Reason is a correct explanation for Assertion
• B. Assertion is True, Reason-2 is True; Reason is Not a correct explanation for Assertion
• C. Assertion is True, Reason is False
• D. Assertion is False, Reason is True

Answer 1

Answer: (D)

Let $A,B,C$ be the respective events of solving the problem and $\overline{A},\overline{B},\overline{C}$ be the respective events of not solving the problem. Then A, B, C are independent event.

$\therefore \overline{A},\overline{B},\overline{C}$are independent events

According to the question, $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3}P\left(C\right)=\frac{1}{4}$

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

Therefore, P( none solves the problem) = P(not A) and (not B) and (not C)

$⟹=P(\overline{A}\cap \overline{B}\cap \overline{C})=P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(\overline{C}\right)$ [as $\overline{A},\overline{B},\overline{C}$are independent events]

$⟹$ P( none solves the problem)=$\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}=\frac{1}{4}$

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

$⟹=1-\frac{1}{4}=\frac{3}{4}\ne \frac{1}{24}$

And also, if $A,B,C$ are three independent events then the probability of at least one of them happening = $1-P(\overline{A}\cap \overline{B}\cap \overline{C})$