A problem is given to three students $A$, $B$, $C$ whose chances of solving it are $\dfrac{1}{2}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$ respectively. If they attempt independently, find the probability that the problem is solved.

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Accepted Answer

The problem is solved if **at least one** of $A,B,C$ solves it. Directly this involves many cases, so use the complement: "solved" is the opposite of "none solves it".

The individual failure probabilities are
$$P(A')=1-	frac12=	frac12,\quad P(B')=1-	frac13=	frac23,\quad P(C')=1-	frac14=	frac34.$$
Since the attempts are independent, so are the failures, and the probability that **all three fail** is the product:
$$P(	ext{none solves})=P(A')P(B')P(C')=\frac12	imes\frac23	imes\frac34=\frac{6}{24}=\frac14.$$
Therefore …

Q.A problem is given to three students $A$ , $B$ , $C$ whose chances of solving it are $\dfrac{1}{2}$ , $\dfrac{1}{3}$ and $\dfrac{1}{4}$ respectively. If they attempt independently, find the probability that the problem is solved.

Concept understanding — Multiplication Theorem and Independent Events

Q.A problem is given to three students AAA, BBB, CCC whose chances of solving it are 12\dfrac{1}{2}21​, 13\dfrac{1}{3}31​ and 14\dfrac{1}{4}41​ respectively. If they attempt independently, find the probability that the problem is solved.

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Q.A problem is given to three students $A$ , $B$ , $C$ whose chances of solving it are $\dfrac{1}{2}$ , $\dfrac{1}{3}$ and $\dfrac{1}{4}$ respectively. If they attempt independently, find the probability that the problem is solved.

Unlock everything free for 14 days