Math JEE Adv. 2025 paper 1

 To find 

$P(T)$, where $T$ is the event that $S_3$ can solve the problem, we'll analyze the given probabilities step-by-step.

Step 1: Define Probabilities

Let:

• $P(S_1)=p_1$,

• $P(S_2)=p_2$,

• $P(S_3)=p_3$.

Step 2: Interpret Given Probabilities

1. Event $U$: At least one of $S_1,S_2,S_3$ can solve the problem.

   $$P(U)=1-(1-p_1)(1-p_2)(1-p_3)=\frac{1}{2}\mathrm{.}$$

   This implies:

   $$(1-p_1)(1-p_2)(1-p_3)=\frac{1}{2}\mathrm{.}$$

2. Event $V$: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve it.

   $$P(V)=\frac{P(S_1\cap \overline{S_2}\cap \overline{S_3})}{P(\overline{S_2}\cap \overline{S_3})}=\frac{p_1(1-p_2)(1-p_3)}{(1-p_2)(1-p_3)}=p_1=\frac{1}{10}\mathrm{.}$$

   So, $p_1=\frac{1}{10}$.

3. Event $W$: $S_2$ can solve the problem and $S_3$ cannot.

   $$P(W)=p_2(1-p_3)=\frac{1}{12}\mathrm{.}$$

Step 3: Solve for $p_2$ and $p_3$

From $P(U)$:

$$(1-p_1)(1-p_2)(1-p_3)=\frac{1}{2}\mathrm{.}$$

Substitute $p_1=\frac{1}{10}$:

$$(1-\frac{1}{10})(1-p_2)(1-p_3)=\frac{1}{2}\Rightarrow \frac{9}{10}(1-p_2)(1-p_3)=\frac{1}{2}\mathrm{.}$$

Simplify:

$$(1-p_2)(1-p_3)=\frac{5}{9}\mathrm{.}$$

From $P(W)$:

$$p_2(1-p_3)=\frac{1}{12}\mathrm{.}$$

Let $q=1-p_3$. Then:

$$(1-p_2)q=\frac{5}{9}\text{and}{p}_{2}q=\frac{1}{12}\mathrm{.}$$

Divide the two equations:

$$\frac{(1-p_2)q}{p_{2}q}=\frac{\frac{5}{9}}{\frac{1}{12}}\Rightarrow \frac{1-p_2}{p_2}=\frac{20}{3}\mathrm{.}$$

Solve for $p_2$:

$$1-p_2=\frac{20}{3}{p}_2\Rightarrow 1=\frac{23}{3}{p}_2\Rightarrow p_2=\frac{3}{23}\mathrm{.}$$

Substitute $p_2$ back into $p_{2}q=\frac{1}{12}$:

$$\frac{3}{23}q=\frac{1}{12}\Rightarrow q=\frac{23}{36}\mathrm{.}$$

Since $q=1-p_3$:

$$1-p_3=\frac{23}{36}\Rightarrow p_3=1-\frac{23}{36}=\frac{13}{36}\mathrm{.}$$

Step 4: Conclusion

The probability $P(T)=p_3=\frac{13}{36}$.

Final Answer: $\boxed{{\displaystyle A}}$