Q.A box contains $3$ orange balls, $3$ green balls and $2$ blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing $2$ green balls and one blue ball is
(A) $\dfrac{3}{28}$
(B) $\dfrac{2}{21}$
(C) $\dfrac{1}{28}$
(D) $\dfrac{167}{168}$

Concept: Conditional Probability (without replacement)

We have 3 green, 3 orange, and 2 blue balls — total 8 balls.

Step 1: Count total ways to draw any 3 balls:

Step 2: Count favourable ways — exactly 2 green and 1 blue:

Choose 2 greens from 3: $\binom{3}{2} = 3$

Choose 1 blue from 2: $\binom{2}{1} = 2$

Favourable combinations: $3 \times 2 = 6$ …

The key idea here is that when drawing without replacement, each ball is equally likely to be chosen at each step. So the probability of a particular colour combination can be found by counting favourable outcomes over total outcomes — either by multiplying conditional probabilities step-by-step, or by using combinations. Both methods give the same result, and we’ll see why.

We have 3 orange, 3 green, and 2 blue balls — 8 balls in total. We want exactly 2 green and 1 blue. Notice that the orange balls are irrelevant to the event; they just fill the rest of the box.

Method 1: Using combinations (faster)

Total number of ways to choose any 3 balls from 8:

Number of ways to choose exactly 2 green from the 3 green balls:

Number of ways to choose exactly 1 blue from the 2 blue balls:

Since the draws are independent in the combinatorial sense (order doesn’t matter), the number of favourable combinations is:

So the probability is:

Method 2: Using conditional probability (step-by-step)

Imagine drawing the three balls one by one without replacement. The event “2 green and 1 blue” can happen in several orders: GGB, GBG, BGG. Each order has the same probability because the draws are symmetric. Let’s compute for one order, say GGB.

Probability first ball is green:

Given that, probability second ball is green (now 2 green left, 7 balls total):

Given that, probability third ball is blue (still 2 blue, 6 balls left):

So for the order GGB: …

Method: A required colour-count when drawing without replacement

Use this when you draw several items at once (or one-by-one without replacement) from a collection of known composition and want a specific breakdown by type.

Steps

Step 1: Total the collection and note the count of each type.

Record how many of each colour/kind there are and the grand total $N$. Items of a type you don't need still count toward $N$.

Step 2: Count total ways to draw the sample (denominator).

Since order does not matter, the number of equally likely selections of $r$ items from $N$ is $\binom{N}{r}$.

Step 3: Count favourable selections (numerator). …

Common Mistakes

Mistake 1: Computing just one order (e.g. green, green, blue) and stopping.

Why it's wrong: $\tfrac38\cdot\tfrac27\cdot\tfrac13 = \tfrac1{28}$ counts only the GGB sequence and gives the distractor $\tfrac1{28}$. Correct approach: multiply by the number of distinct orders (GGB, GBG, BGG), or use combinations which sidestep order entirely: $\dfrac{\binom{3}{2}\binom{2}{1}}{\binom{8}{3}}=\dfrac{6}{56}=\dfrac{3}{28}$.

Mistake 2: Forgetting the orange balls still count in the total. …