Probability — Conditional Probability & Independence
DAILY_CHALLENGE
Grade None
Question:
Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_1$ be the event that the ball chosen belonged to Box I and let $E_2$ be the event that the ball chosen belonged to Box II. Let $F_1$ be the event that the ball chosen is red and let $F_2$ be the event that the ball chosen is green.
Then which of the following statements is (are) TRUE?
The events $E_1$ and $F_1$ are independent
The events $E_2$ and $F_2$ are dependent
The conditional probability $P(F_1 \mid E_1)$ is equal to the conditional probability $P(F_1 \mid E_2)$
The conditional probability $P(F_1 \mid E_1)$ is greater than the conditional probability $P(F_2 \mid E_2)$
Step-by-Step Solution
Key Concept: The red fraction in Box I ($6/15 = 2/5$) equals the overall red fraction ($14/35 = 2/5$) because Box II also has a $2/5$ red fraction. This equality is precisely what makes box-membership and colour independent.
**Step 1: Compute base probabilities**
Box I: 6R + 9G = 15 balls. Box II: 8R + 12G = 20 balls. Total: 35 balls.
$P(E_1)=\tfrac{15}{35}=\tfrac{3}{7}$, $P(E_2)=\tfrac{4}{7}$, $P(F_1)=\tfrac{14}{35}=\tfrac{2}{5}$, $P(F_2)=\tfrac{21}{35}=\tfrac{3}{5}$.
**Step 2: Check (A): independence of $E_1$ and $F_1$**
$P(E_1 \cap F_1) = \tfrac{6}{35}$. $P(E_1)\cdot P(F_1) = \tfrac{3}{7}\cdot\tfrac{2}{5} = \tfrac{6}{35}$. Equal ⟹ independent. ✓
**Step 3: Check (B): dependence of $E_2$ and $F_2$**
$P(E_2 \cap F_2) = \tfrac{12}{35}$. $P(E_2)\cdot P(F_2) = \tfrac{4}{7}\cdot\tfrac{3}{5} = \tfrac{12}{35}$. Equal ⟹ independent, NOT dependent. ✗
**Step 4: Check (C): compare $P(F_1|E_1)$ and $P(F_1|E_2)$**
$P(F_1|E_1) = \tfrac{6}{15} = \tfrac{2}{5}$. $P(F_1|E_2) = \tfrac{8}{20} = \tfrac{2}{5}$. Equal. ✓
**Step 5: Check (D): compare $P(F_1|E_1)$ and $P(F_2|E_2)$**
$P(F_1|E_1) = \tfrac{2}{5} = 0.4$. $P(F_2|E_2) = \tfrac{12}{20} = 0.6$. So $P(F_1|E_1) < P(F_2|E_2)$. ✗
Correct Answer: A, C