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, $P(S)\in \left(\frac{2}{3},1\right)$