Q.A bag contains $5$ red and $3$ blue balls. If $3$ balls are drawn at random without replacement, the probability of getting exactly one red ball is
(A) $\dfrac{45}{196}$
(B) $\dfrac{135}{392}$
(C) $\dfrac{15}{56}$
(D) $\dfrac{15}{29}$

Concept: Conditional Probability (without replacement)

We want exactly 1 red ball in 3 draws without replacement. That means 1 red and 2 blue balls.

Step 1: Total balls = $5$ red + $3$ blue = $8$.

Number of ways to choose any 3 balls: $\binom{8}{3} = 56$.

Step 2: Number of ways to choose exactly 1 red (from 5) and 2 blue (from 3): …

We are dealing with conditional probability without replacement — each draw changes the composition of the bag. The key idea: "exactly one red" means we get 1 red and 2 blue balls, in any order. Since the draws are without replacement, the probability is not constant across draws; we must account for the changing counts.

A clean way: count the number of favorable combinations and divide by the total number of ways to choose 3 balls from 8. This avoids the messy multiplication of conditional probabilities for each order.

Here, successes = red balls (5), failures = blue balls (3), $n = 3$, $k = 1$.

1. Total number of ways to choose any 3 balls from 8

   This is $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

2. Number of ways to choose exactly 1 red ball from the 5 red balls

   That’s $\binom{5}{1} = 5$.

3. Number of ways to choose the remaining 2 balls from the 3 blue balls

   That’s $\binom{3}{2} = 3$.

4. Number of favorable combinations

   Multiply the independent choices: $5 \times 3 = 15$.

5. Probability …

Method: "Exactly $k$" without replacement via combinations

Use this counting method whenever balls/cards are drawn without replacement and you want exactly $k$ of one type.

Steps

Step 1: Count the total unordered selections.

Choosing $n$ items from $N$: $\dbinom{N}{n}$ — this is the denominator (all equally likely selections).

Step 2: Count the favourable selections.

Choose $k$ from the target group and the rest from the others:

Step 3: Form the ratio (hypergeometric probability). …

Common Mistakes

Mistake 1: Using the binomial (with-replacement) formula.

Why it's wrong: the balls are drawn without replacement, so the probability changes after each draw; the binomial answer $\binom{3}{1}(5/8)(3/8)^2$ isn't even among the options. Correct approach: use combinations (hypergeometric counting).

Mistake 2: Forgetting the different orders — or double-counting them. …