A problem is given to three students $A,B$ and $C$, whose probabilities of solving the problem independently are $\frac{1}{2},\frac{3}{4}$ and $p$ respectively. If the probability that the problem can be solved is $\frac{29}{32}$, then what is the value of $p$?

Question

YESPYQ · Accepted Answer

Correct answer: d) $\frac{1}{4}$. Given that, $P(A)=\frac{1}{2},P(B)=\frac{3}{4}$ and $P(C)=p$ Probability that the problem can not be solved $=P(\overset{¯}{A})⋅P(\overset{¯}{B})⋅P(\overset{¯}{C})$ $=\left(1−\frac{1}{2}\right)\left(1−\frac{3}{4}\right)(1−p)$ $=\frac{1}{2}\times \frac{1}{4}(1−p)=\frac{1−p}{8}$ $∴$ Probability that the problem can be solved $=1-$Probability that the problem cannot be solved $\Rightarrow \frac{29}{32}=1−\frac{(1−p)}{8}$ $\Rightarrow 1−p=\frac{3}{4}$ $∴p=\frac{1}{4}$

A problem is given to three students \(A,B\) and \(C\), whose probabilities of solving the problem independently are \(\…

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