Q.A die is thrown and a card is selected at random from a deck of $52$ playing cards. The probability of getting an even number on the die and a spade card is
(A) $\dfrac{1}{2}$
(B) $\dfrac{1}{4}$
(C) $\dfrac{1}{8}$
(D) $\dfrac{3}{4}$

Concept: Conditional Probability — Since the die throw and card draw are independent events, the joint probability is the product of their individual probabilities.

Step 1: Probability of an even number on a die.

A standard die has 6 outcomes: 2, 4, 6 are even.

$P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.

Step 2: Probability of drawing a spade from a deck of 52 cards.

There are 13 spades.

$P(\text{spade}) = \frac{13}{52} = \frac{1}{4}$. …

The core idea here is conditional probability — but in its simplest form: independence. When two events do not influence each other, the probability that both occur is just the product of their separate probabilities. That’s exactly the situation when you throw a die and draw a card: the outcome of the die has no effect on which card you pick, and vice versa.

Let’s break it down.

1. Probability of an even number on the die A standard die has six faces: $1, 2, 3, 4, 5, 6$. The even numbers among these are $2, 4, 6$ — that’s 3 outcomes. So

2. Probability of drawing a spade from a deck of 52 cards A standard deck has 4 suits (spades, hearts, diamonds, clubs), each with 13 cards. So there are 13 spades. Hence

3. Since the die throw and card draw are independent, the probability that both happen is $P(\text{even and spade}) = P(\text{even}) \times P(\text{spade}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}.$ …

Method: Probability of "$A$ and $B$" for two independent experiments

Use this when two separate random experiments (a die and a card, two coins, a spinner and a draw) run without influencing each other and you want the probability that a chosen result happens in both.

Steps

Step 1: Confirm the two experiments are independent.

One experiment's outcome must not change the other's odds. Physically separate devices — a die and a deck of cards — are independent by default.

Step 2: Find each event's probability on its own. …

Common Mistakes

Mistake 1: Adding the two probabilities instead of multiplying, $\tfrac12 + \tfrac14 = \tfrac34$.

Why it's wrong: addition applies to "$A$ or $B$" (for mutually exclusive events), not to "$A$ and $B$." Correct approach: for independent events happening together, multiply: $\tfrac12\times\tfrac14 = \tfrac18$. …