Q.A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

The key idea is Hypergeometric Probability — drawing without replacement from a finite population of two types.

We need the probability that all 3 drawn oranges are good.

Step 1: Total ways to choose 3 oranges from 15:

$\binom{15}{3}$

Step 2: Favorable ways — choose 3 good oranges from the 12 good ones:

$\binom{12}{3}$

Step 3: Probability = favorable / total:

Why Hypergeometric Probability?

When you draw items without replacement from a finite population that has two distinct types (here: good vs. bad oranges), the probability of getting a certain number of "successes" (good oranges) follows the hypergeometric distribution.

The key difference from the binomial distribution: because we don't replace the oranges, the probability of picking a good orange changes after each draw. The hypergeometric formula handles this by counting combinations directly.

Here, "success" = good orange, "failure" = bad orange, $n$ = number drawn.

Step-by-step solution

1. Identify the numbers

• Total oranges: $N = 15$
• Good oranges (successes in population): $K = 12$
• Bad oranges (failures): $N - K = 3$
• Oranges drawn: $n = 3$
• We need all 3 drawn to be good, so $k = 3$ successes.

2. Apply the hypergeometric formula

We want:

3. Compute each combination

• $\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$
• $\binom{3}{0} = 1$ (there's exactly one way to choose zero bad oranges)
• $\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$

4. Put it together

5. Simplify the fraction

Divide numerator and denominator by 5:

Method: Probability That Every Item Drawn Without Replacement Is of One Type

Use this whenever a sample is drawn without replacement from a finite group of two kinds and you need the chance that all drawn items are the "good" kind.

Steps

Step 1: Set up the dependent chain

Drawing without replacement makes the draws dependent, so use the multiplication theorem, updating the counts each time:

where $g$ is the number of good items and $N$ the total, each reduced by one per draw.

Step 2: Or count equally likely selections

Equivalently, since all selections are equally likely, take favourable over total using combinations:

Step 3: Simplify and sanity-check

Reduce the resulting fraction. Note this is not a binomial/constant-probability situation — because there is no replacement the per-draw probability changes, so never raise a single probability to a power here.

Common Mistakes

Mistake 1: Using a binomial (with-replacement) probability.

Students take a constant $p=\frac{12}{15}$ and compute $\left(\frac{12}{15}\right)^3=\frac{64}{125}$. Why it's wrong: the oranges are drawn without replacement, so the fraction of good ones changes after each pick. Correct approach: $\frac{12}{15}\times\frac{11}{14}\times\frac{10}{13}=\frac{44}{91}$, or $\dfrac{\binom{12}{3}}{\binom{15}{3}}$.

Mistake 2: Not reducing the counts after each draw.

Writing $\frac{12}{15}\times\frac{12}{14}\times\frac{12}{13}$ keeps 12 good oranges every time. Why it's wrong: each good orange removed drops both the good count and the total by one. Correct approach: $12/15,\ 11/14,\ 10/13$.