Q.Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

Concept: Hypergeometric Probability (with replacement → Binomial trials)

Since each draw is independent and the composition of the box stays the same, we treat each draw as a Bernoulli trial.

Step 1 – Total balls

$10$ black + $8$ red = $18$ balls.

Step 2 – Probability of a red in one draw

$P(R) = \frac{8}{18} = \frac{4}{9}$; similarly $P(B) = \frac{10}{18} = \frac{5}{9}$.

Step 3 – Compute each case

• (i) Both red: $P(RR) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$
• (ii) First black, second red: $P(BR) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$ …

The key insight here is the phrase "with replacement". Many students instinctively reach for the hypergeometric formula (which is for without replacement), but that would be wrong. With replacement, the box is reset after each draw — the probability of drawing a red ball is the same on the second draw as it was on the first.

So this is a binomial (or simply independent-events) problem. Let's break it down.

Step-by-step solution

1. Find the total number of balls and the individual probabilities.

Total balls = $10$ black + $8$ red = $18$.

Probability of drawing a red ball on any single draw:

Probability of drawing a black ball on any single draw:

Because we replace the ball, these probabilities are the same for every draw.

2. (i) Both balls are red.

We need $P(\text{first is red AND second is red})$. Since draws are independent:

3. (ii) First ball is black and second is red.

Again, independent events:

4. (iii) One of them is black and the other is red. …

Method: Drawing with replacement — independent draws multiply

Use this for repeated draws with replacement (the box/deck is restored each time), so every draw has the same, unchanging probabilities and the draws are independent.

Steps

Step 1: Fix the single-draw probabilities.

With replacement the composition never changes, so each colour/type keeps a constant probability, e.g. $P(\text{red}) = \dfrac{\#\text{red}}{\#\text{total}}$ on every draw.

Step 2: For a specific ordered outcome, multiply.

Because the draws are independent,

Step 3: For an unordered outcome, add the mutually exclusive orders. …

Common Mistakes

Mistake 1: Treating the draws as without replacement and changing the counts (e.g. denominator $17$ on the second draw).

Why it's wrong: the ball is replaced, so the box is reset and each draw keeps the same probabilities out of $18$. Correct approach: use constant $P(R)=\tfrac49$, $P(B)=\tfrac59$ on every draw. …