Q.Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Concept: Conditional Probability — the probability of the second event depends on the outcome of the first, since there is no replacement.

Step 1: In a standard deck, there are 26 black cards out of 52.

Probability the first card is black:

Step 2: After drawing one black card, 25 black cards remain out of 51 total cards.

Probability the second card is black, given the first was black:

Step 3: Multiply the probabilities (since both events must occur):

Why conditional probability is the natural tool here

When we draw cards without replacement, the outcome of the second draw depends on what happened in the first draw. This is the classic setting for conditional probability: we want $P(\text{both black}) = P(\text{first black}) \times P(\text{second black} \mid \text{first black})$.

The intuition is simple. After one black card is removed, the deck has fewer black cards and fewer total cards. The probability for the second draw must reflect that changed situation. Multiplying the two probabilities along the "path" of the event gives the joint probability.

Let's apply this step by step.

1. Probability that the first card is black

   A standard deck has 52 cards, of which 26 are black (spades and clubs). So:

2. Probability that the second card is black, given the first was black

   After removing one black card, the deck now has 51 cards left, and only 25 of them are black. So the conditional probability is:

3. Multiply to get the joint probability

This fraction is already in its simplest form (25 and 102 share no common factor other than 1).

Method: Probability of a Sequence of Draws Without Replacement

Use this whenever items are drawn one after another without replacement and you want the probability that they are all of a specified type (all black, all defective, etc.).

Steps

Step 1: Chain the draws with the multiplication theorem

Because each draw changes what is left, the events are dependent. The multiplication theorem for dependent events chains conditional probabilities:

Step 2: Update both counts after each draw

For every successive draw, reduce the favourable count in the numerator and the total count in the denominator by the items already removed. Forgetting to shrink the total is the commonest error.

Step 3: Multiply the chain (or count with combinations)

Multiply the conditional probabilities along the path. As a cross-check you may instead count equally likely selections, $\dfrac{\binom{\text{favourable}}{k}}{\binom{\text{total}}{k}}$ — both routes give the same answer.

Common Mistakes

Mistake 1: Keeping the deck unchanged on the second draw.

Students write $\frac{26}{52}\times\frac{26}{51}$, holding the black count at 26. Why it's wrong: the draw is without replacement, so after one black card leaves, only 25 black cards remain among 51. Correct approach: $\frac{26}{52}\times\frac{25}{51}=\frac{25}{102}$.

Mistake 2: Changing only the total, not the favourable count.

Some reduce the denominator to 51 but keep 26 on top. Why it's wrong: both the black count and the total fall by one after the first black card. Correct approach: numerator $26\to25$ and denominator $52\to51$.

Mistake 3: Treating the two draws as independent.

Multiplying $\frac12\times\frac12$ assumes replacement. Why it's wrong: the second draw is conditional on the first. Correct approach: use $P(\text{both})=P(\text{1st black})\,P(\text{2nd black}\mid\text{1st black})$.