A problem in mathematics is given to $3$ students whose chances of solving it are $ \frac13,\frac14,\frac15 $. What is the probability that the problem is solved?
Shouldn't the probability of the problem being solved be $\frac13×\frac14×\frac15=\frac1{60}$. But the answer given is $\frac35$. Please explain why I am wrong?
You calculated the probability that all three solve it.
The probability that none solve it is $\frac23×\frac34×\frac45=\frac25$, and the probability that at least one does (the question asks for this) is the complement, or $\frac35$.
Let $A,B,C$ be the events of solving the problem and $\overline{A},\overline{B},\overline{C}$ be the events of not solving the problem.
We have $P(A)=\dfrac13,P(B)=\dfrac14,P(C)=\dfrac15$
$P(\overline{A})=\dfrac23,P(\overline{B})=\dfrac34,P(\overline{C})=\dfrac45$
$P(\mbox{ none solves the problem})=P(\overline{A}\cap\overline{B}\cap\overline{C})$
$=\dfrac23\times\dfrac34\times\dfrac45=\dfrac25$
Therefore,
$P(\mbox{ the problem will be solved})=1-P(\mbox{ none solves the problem})$
$$1-\dfrac25=\dfrac35$$
