Q.One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events $E$ and $F$ independent?

Concept: Event Independence — Two events $E$ and $F$ are independent iff $P(E \cap F) = P(E) \cdot P(F)$.

(i) $E$: spade, $F$: ace.

$P(E) = \frac{13}{52} = \frac14$, $P(F) = \frac{4}{52} = \frac1{13}$, $P(E \cap F) = P(\text{ace of spades}) = \frac1{52}$.

Check: $\frac14 \cdot \frac1{13} = \frac1{52}$. They are equal → independent.

(ii) $E$: black, $F$: king.

$P(E) = \frac{26}{52} = \frac12$, $P(F) = \frac{4}{52} = \frac1{13}$, $P(E \cap F) = P(\text{black king}) = \frac{2}{52} = \frac1{26}$.

Check: $\frac12 \cdot \frac1{13} = \frac1{26}$. Equal → independent.

(iii) $E$: king or queen, $F$: queen or jack. …

The core idea: independence means knowing one event happened gives you no information about whether the other happened. Mathematically, we check whether the probability of both happening together equals the product of their individual probabilities. In a deck of cards, this often works when the events are about different "categories" (like suit and rank) that don't overlap in a special way.

Let's go through each case.

Case (i): $E$ = spade, $F$ = ace

1. Find $P(E)$: There are 13 spades in 52 cards, so $P(E) = \frac{13}{52} = \frac{1}{4}$.

2. Find $P(F)$: There are 4 aces, so $P(F) = \frac{4}{52} = \frac{1}{13}$.

3. Find $P(E \cap F)$: The card that is both a spade and an ace is the ace of spades — exactly 1 card. So $P(E \cap F) = \frac{1}{52}$.

4. Check independence:

   $P(E) \cdot P(F) = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}$.

   This equals $P(E \cap F)$. So E and F are independent.

Case (ii): $E$ = black, $F$ = king

1. $P(E)$: Half the deck is black (clubs + spades), so $P(E) = \frac{26}{52} = \frac{1}{2}$.

2. $P(F)$: 4 kings, so $P(F) = \frac{4}{52} = \frac{1}{13}$.

3. $P(E \cap F)$: Black kings — there are 2 (king of clubs and king of spades). So $P(E \cap F) = \frac{2}{52} = \frac{1}{26}$.

4. Check:

   $P(E) \cdot P(F) = \frac{1}{2} \cdot \frac{1}{13} = \frac{1}{26}$.

   Matches $P(E \cap F)$. So E and F are independent.

Case (iii): $E$ = king or queen, $F$ = queen or jack

1. $P(E)$: There are 4 kings + 4 queens = 8 cards. But careful — no overlap between kings and queens, so $P(E) = \frac{8}{52} = \frac{2}{13}$. …

Method: The product test for independence

For any "are these two events independent?" question — cards, dice, or given probabilities — the decision rests on a single equality, checked case by case.

Steps

Step 1: Compute the two individual probabilities.

Count favourable outcomes over the sample space to get $P(E)$ and $P(F)$.

Step 2: Compute the joint probability $P(E \cap F)$ directly.

Count the outcomes that satisfy both descriptions at once (e.g. a card that is both a spade and an ace) — do not multiply here; find the actual overlap.

Step 3: Apply the test. …

Common Mistakes

Mistake 1: Assuming that because $E$ and $F$ share a rank (e.g. "queen" lies in both sets of case iii) they must be dependent — or independent — without testing.

Why it's wrong: independence is decided only by $P(E\cap F)=P(E)P(F)$, not by whether the descriptions overlap in words. Correct approach: compute and compare for each case; case (iii) fails the test.

Mistake 2: Mis-counting the overlap cards. …