Q.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

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

Step 1: Total balls = $10 + 5 = 15$.

Probability that the first ball is black:

Step 2: After removing one black ball, remaining balls = $14$, black balls left = $9$.

Probability that the second ball is black, given the first was black:

Step 3: Multiply the probabilities (chain rule for conditional probability):

Why conditional probability works here

When we draw without replacement, the outcome of the first draw changes the composition of the urn for the second draw. That’s the heart of conditional probability: we want $P(\text{first black AND second black})$, which we can write as:

This is not just a formula — it’s common sense. If the first ball is black, the urn now has 9 black and 5 white balls left. The second draw’s probability depends entirely on what happened first.

Step-by-step solution

1. Probability that the first ball is black

Total balls initially: $10 + 5 = 15$.

Black balls: $10$.

So:

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

After removing one black ball, the urn has:

• Black balls left: $10 - 1 = 9$
• Total balls left: $15 - 1 = 14$

Thus:

3. Multiply the two probabilities

Method: The multiplication theorem for draws without replacement

Use this whenever objects are drawn one after another without replacement and you want the probability that a specific sequence occurs, because each draw changes what remains.

Steps

Step 1: Write the joint event as a chain of conditionals.

The second factor is conditional because removing the first object alters the pool.

Step 2: Compute each factor by updating the counts.

Find $P(A)$ from the original composition, then recompute the pool for $P(B\mid A)$: subtract one from both the favourable count and the total (e.g. $\tfrac{10}{15}$ then $\tfrac{9}{14}$).

Step 3: Multiply.

Multiply the successive probabilities. As a check, the combinations method $\binom{\text{favourable}}{k}/\binom{\text{total}}{k}$ gives the same answer when order does not matter.

Common Mistakes

Mistake 1: Treating the two draws as independent (with replacement).

Why it's wrong: without replacement the urn shrinks, so the second probability is $\tfrac{9}{14}$, not $\tfrac{10}{15}$ again; using $\tfrac{10}{15}\times\tfrac{10}{15}$ gives the wrong $\tfrac49$. Correct approach: use $P(\text{both black})=\tfrac{10}{15}\times\tfrac{9}{14}=\tfrac37$.

Mistake 2: Forgetting to reduce the total from $15$ to $14$ for the second draw.

Why it's wrong: one ball has already been removed, so both the black count and the total drop by one. Correct approach: update numerator and denominator together at each step.