Question 2

Let $A$, $B$, and $C$ be the events that students $S_1$, $S_2$, and $S_3$ solve the problem, respectively.

Given:

• $P(U) = P(A \cup B \cup C) = \frac{1}{2}$
• $P(V) = P(A \mid B^c \cap C^c) = \frac{1}{10}$
• $P(W) = P(B \cap C^c) = \frac{1}{12}$

We need to find $P(T) = P(C)$.

Step 1: Find the probability that no one solves the problem. The complement of "at least one solves it" is "no one solves it" ($A^c \cap B^c \cap C^c$).

$P(A^c \cap B^c \cap C^c) = 1 - P(A \cup B \cup C) = 1 - \frac{1}{2} = \frac{1}{2}$

Step 2: Use the conditional probability formula for event V. By definition of conditional probability:

$P(A \mid B^c \cap C^c) = \frac{P(A \cap B^c \cap C^c)}{P(B^c \cap C^c)} = \frac{1}{10}$

Step 3: Relate the intersection probabilities to find $P(B^c \cap C^c)$. The event that "neither $S_2$ nor $S_3$ solves the problem" ($B^c \cap C^c$) can be split into two mutually exclusive scenarios: either $S_1$ solves it, or $S_1$ doesn't solve it.

$P(B^c \cap C^c) = P(A \cap B^c \cap C^c) + P(A^c \cap B^c \cap C^c)$

From Step 2, we know $P(A \cap B^c \cap C^c) = \frac{1}{10} P(B^c \cap C^c)$.

Substituting this and the value from Step 1 into our equation:

$P(B^c \cap C^c) = \frac{1}{10} P(B^c \cap C^c) + \frac{1}{2}$

$\frac{9}{10} P(B^c \cap C^c) = \frac{1}{2}$

$P(B^c \cap C^c) = \frac{5}{9}$

Step 4: Find the probability that $S_3$ does not solve it ($P(C^c)$). The event that "$S_3$ does not solve the problem" ($C^c$) can be split into two scenarios: either $S_2$ solves it, or $S_2$ doesn't solve it.

$P(C^c) = P(B \cap C^c) + P(B^c \cap C^c)$

We are given $P(B \cap C^c) = \frac{1}{12}$, and we just found $P(B^c \cap C^c) = \frac{5}{9}$.

$P(C^c) = \frac{1}{12} + \frac{5}{9} = \frac{3}{36} + \frac{20}{36} = \frac{23}{36}$

Step 5: Calculate the final probability $P(T)$.

$P(T) = P(C) = 1 - P(C^c)$

$P(T) = 1 - \frac{23}{36} = \frac{13}{36}$

Answer: $\frac{13}{36}$