Question

Three students $S_1, S_2$, and $S_3$ are given a problem to solve. Consider the following events:

$U$: At least one of $S_1, S_2$, and $S_3$ can solve the problem,

$V$: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem,

$W$: $S_2$ can solve the problem and $S_3$ cannot solve the problem,

$T$: $S_3$ can solve the problem.

For any event $E$, let $P(E)$ denote the probability of $E$. If

$P(U)=\frac{1}{2}$, $P(V)=\frac{1}{10}$, and $P(W)=\frac{1}{12}$,

then $P(T)$ is equal to

• A. $\frac{13}{36}$
• B. $\frac{1}{3}$
• C. $\frac{19}{60}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{13}{36}$

Answer 2

Let $S_1, S_2, S_3$ be the events that students $S_1, S_2, S_3$ can solve the problem, respectively. Let $S_i^c$ be the event that student $S_i$ cannot solve the problem.

The event $U$ is "At least one of $S_1, S_2$, and $S_3$ can solve the problem". This is $S_1 \cup S_2 \cup S_3$. We are given $P(U) = P(S_1 \cup S_2 \cup S_3) = 1/2$. The complement of $U$, $U^c$, is the event that none of them can solve the problem, which is $S_1^c \cap S_2^c \cap S_3^c$. $P(U^c) = P(S_1^c \cap S_2^c \cap S_3^c) = 1 - P(U) = 1 - 1/2 = 1/2$.

The event $V$ is "$S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem". This is the conditional probability $P(S_1 | S_2^c \cap S_3^c)$. We are given $P(V) = P(S_1 | S_2^c \cap S_3^c) = 1/10$. Using the definition of conditional probability, $P(S_1 | S_2^c \cap S_3^c) = \frac{P(S_1 \cap S_2^c \cap S_3^c)}{P(S_2^c \cap S_3^c)}$. So, $\frac{P(S_1 \cap S_2^c \cap S_3^c)}{P(S_2^c \cap S_3^c)} = 1/10$.

Consider the event $S_2^c \cap S_3^c$. This event can be partitioned into two disjoint events based on whether $S_1$ solves the problem or not: $(S_1 \cap S_2^c \cap S_3^c)$ and $(S_1^c \cap S_2^c \cap S_3^c)$. So, $P(S_2^c \cap S_3^c) = P(S_1 \cap S_2^c \cap S_3^c) + P(S_1^c \cap S_2^c \cap S_3^c)$. Let $a = P(S_1 \cap S_2^c \cap S_3^c)$. We know $P(S_1^c \cap S_2^c \cap S_3^c) = 1/2$. So, $P(S_2^c \cap S_3^c) = a + 1/2$. Substitute this into the equation from $P(V)$: $\frac{a}{a + 1/2} = 1/10$. $10a = a + 1/2$ $9a = 1/2$ $a = 1/18$. So, $P(S_1 \cap S_2^c \cap S_3^c) = 1/18$. And $P(S_2^c \cap S_3^c) = 1/18 + 1/2 = 1/18 + 9/18 = 10/18 = 5/9$.

The event $W$ is "$S_2$ can solve the problem and $S_3$ cannot solve the problem". This is $S_2 \cap S_3^c$. We are given $P(W) = P(S_2 \cap S_3^c) = 1/12$.

The event $T$ is "$S_3$ can solve the problem". This is $S_3$. We want to find $P(T) = P(S_3)$. The complement of $T$ is $T^c = S_3^c$. We have $P(S_3) = 1 - P(S_3^c)$.

Consider the event $S_3^c$. This event can be partitioned based on the outcomes of $S_1$ and $S_2$: $S_3^c = (S_1 \cap S_2 \cap S_3^c) \cup (S_1 \cap S_2^c \cap S_3^c) \cup (S_1^c \cap S_2 \cap S_3^c) \cup (S_1^c \cap S_2^c \cap S_3^c)$. These four events are mutually exclusive. $P(S_3^c) = P(S_1 \cap S_2 \cap S_3^c) + P(S_1 \cap S_2^c \cap S_3^c) + P(S_1^c \cap S_2 \cap S_3^c) + P(S_1^c \cap S_2^c \cap S_3^c)$.

We know: $P(S_1 \cap S_2^c \cap S_3^c) = 1/18$. $P(S_1^c \cap S_2^c \cap S_3^c) = 1/2$.

Consider the event $S_2 \cap S_3^c$. This event is the union of two disjoint events: $S_2 \cap S_3^c = (S_1 \cap S_2 \cap S_3^c) \cup (S_1^c \cap S_2 \cap S_3^c)$. So, $P(S_2 \cap S_3^c) = P(S_1 \cap S_2 \cap S_3^c) + P(S_1^c \cap S_2 \cap S_3^c)$. We are given $P(S_2 \cap S_3^c) = 1/12$. So, $P(S_1 \cap S_2 \cap S_3^c) + P(S_1^c \cap S_2 \cap S_3^c) = 1/12$.

Now substitute the known probabilities into the expression for $P(S_3^c)$: $P(S_3^c) = [P(S_1 \cap S_2 \cap S_3^c) + P(S_1^c \cap S_2 \cap S_3^c)] + P(S_1 \cap S_2^c \cap S_3^c) + P(S_1^c \cap S_2^c \cap S_3^c)$. $P(S_3^c) = P(S_2 \cap S_3^c) + P(S_1 \cap S_2^c \cap S_3^c) + P(S_1^c \cap S_2^c \cap S_3^c)$. $P(S_3^c) = 1/12 + 1/18 + 1/2$.

To sum these fractions, find a common denominator, which is 36. $1/12 = 3/36$. $1/18 = 2/36$. $1/2 = 18/36$. $P(S_3^c) = 3/36 + 2/36 + 18/36 = (3 + 2 + 18)/36 = 23/36$.

Finally, $P(T) = P(S_3) = 1 - P(S_3^c) = 1 - 23/36 = (36 - 23)/36 = 13/36$.

This matches option (A). The solution did not require assuming independence of the events $S_1, S_2, S_3$.