Question

The probability of a problem being solved by $3$ students independently are $\frac{1}{2}, \, \frac{1}{3}$ and $\alpha $ respectively. If the probability that the problem is solved is $P\left(S\right)$ , then $P\left(S\right)$ lies in the interval (where, $\alpha \in \left(0 , 1\right)$ )

• A. $\left(0 , \frac{1}{2}\right)$
• B. $\left(\frac{1}{3} , \frac{1}{2}\right)$
• C. $\left(\frac{2}{3} , 1\right)$
• D. $\left(\frac{1}{3} , \frac{2}{3}\right)$

Answer 1

Answer: (C)

Let 3 students are $A, B, C$ and their probability of solving problems is $P(A), P(B), P(C)$ respectively.
$P(A)=\frac{1}{2}, P(B)=\frac{1}{3}, P(C)=\alpha$
Probability of the problem is being solved, $P(S)=1-P\left(A^{C} \cap B^{C} \cap C^{C}\right)$
$=1-P\left(A^{C}\right) P\left(B^{C}\right) P\left(C^{C}\right)$
$=1-(1-P(A))(1-P(B))(1-P(C))$
$=1-\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)(1-\alpha)$
$=\frac{2}{3}+\frac{1}{3} \alpha$
Hence, $P(S) \in\left(\frac{2}{3}, 1\right)$