Q.A fair die is rolled. Consider events $E = \{1,3,5\}$, $F = \{2,3\}$ and $G = \{2,3,4,5\}$ Find

Concept: Conditional Probability — $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) \neq 0$.

Step 1: List probabilities.

Since the die is fair, each outcome has probability $\frac{1}{6}$.

$P(E) = \frac{3}{6} = \frac{1}{2}$, $P(F) = \frac{2}{6} = \frac{1}{3}$, $P(G) = \frac{4}{6} = \frac{2}{3}$.

Step 2: Find intersections.

$E \cap F = \{3\}$ → $P(E \cap F) = \frac{1}{6}$.

$E \cap G = \{3,5\}$ → $P(E \cap G) = \frac{2}{6} = \frac{1}{3}$.

$F \cap G = \{2,3\}$ → $P(F \cap G) = \frac{2}{6} = \frac{1}{3}$.

$E \cup F = \{1,2,3,5\}$ → $P(E \cup F) = \frac{4}{6} = \frac{2}{3}$.

$(E \cup F) \cap G = \{2,3,5\}$ → $P((E \cup F) \cap G) = \frac{3}{6} = \frac{1}{2}$.

$(E \cap F) \cap G = \{3\}$ → $P((E \cap F) \cap G) = \frac{1}{6}$.

Step 3: Apply formula.

(i) $P(E|F) = \frac{1/6}{1/3} = \frac{1}{2}$, $P(F|E) = \frac{1/6}{1/2} = \frac{1}{3}$. …

Each outcome of $\{1,2,3,4,5,6\}$ has probability $\tfrac16$. Given $E=\{1,3,5\}$, $F=\{2,3\}$, $G=\{2,3,4,5\}$:

(i) $P(E\mid F)$ and $P(F\mid E)$. $E\cap F=\{3\}$, so $P(E\cap F)=\tfrac16$.

(ii) $P(E\mid G)$ and $P(G\mid E)$. $E\cap G=\{3,5\}$, so $P(E\cap G)=\tfrac26=\tfrac13$.

$P(E\mid G)=\frac{1/3}{2/3}=\frac12,\qquad P(G\mid E)=\frac{1/3}{1/2}=\frac23.$ …

Method: Computing conditional probabilities directly from given event sets

Use this when the events are handed to you as explicit subsets of the sample space and you must evaluate several conditionals, including compound events $E\cup F$ and $E\cap F$, and both directions $P(A\mid B)$ and $P(B\mid A)$.

Steps

Step 1: Fix the outcome probabilities.

For a fair die each of $\{1,\dots,6\}$ has probability $\tfrac16$, so any event's probability is (its size)$/6$. Record $P(E),P(F),P(G)$ once.

Step 2: Build the intersections you will need.

Work out the needed overlaps as sets first — $E\cap F$, $E\cap G$, and for compound conditionals $(E\cup F)\cap G$ and $(E\cap F)\cap G$. Use $E\cup F$ = outcomes in either, $E\cap F$ = outcomes in both.

Step 3: Apply the definition each time.

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$ …

Common Mistakes

Mistake 1: Swapping $P(E|F)$ with $P(F|E)$.

Why it's wrong: both share the numerator $P(E\cap F)=\frac16$ but divide by different quantities — $P(E|F)=\frac{1/6}{1/3}=\frac12$ while $P(F|E)=\frac{1/6}{1/2}=\frac13$. Correct approach: always divide by the probability of the event written after the bar.

Mistake 2: Forgetting to intersect with $G$ before dividing in part (iii). …