Q.A problem in Mathematics is given to three students whose chances of solving it are $\dfrac{1}{3}, \dfrac{1}{4}$, and $\dfrac{1}{5}$.

Step 1. Given. $P_1=\dfrac13$, $P_2=\dfrac14$, $P_3=\dfrac15$ -- the (independent) chances each student solves the problem, so $P(\bar1)=\dfrac23$, $P(\bar2)=\dfrac34$, $P(\bar3)=\dfrac45$.

Step 2. Part (i) -- probability the problem is solved. The problem stays unsolved only if ALL THREE fail: $P(\text{none solve})=\dfrac23\times\dfrac34\times\dfrac45=\dfrac{2\times3\times4}{3\times4\times5}=\dfrac{24}{60}=\dfrac25$. So $P(\text{solved})=1-\dfrac25=\dfrac35$.

Step 3. Part (ii) -- exactly one solves. Sum the three 'this one solves, the other two fail' terms:

1 solves, 2&3 fail: $\dfrac13\times\dfrac34\times\dfrac45=\dfrac{12}{60}=\dfrac15$. …

Independence across the three students; 'at least one' via the complement of 'none'; 'exactly one' …

• Adding $P_1+P_2+P_3$ directly for part (i) instead of using $1-P(\text{none})$ (that sum exceeds 1 and double-counts overlapping success cases). …