Two Successes (Without Replacement)

Statement

When drawing without replacement, the probability of two successes changes after the first draw. The formula is:

\[ P = \frac{k}{n} \times \frac{k-1}{n-1} \]

Here, \(k\) is the number of favourable outcomes, and \(n\) is the total outcomes.

Why it’s true

• The probability of success on the first draw is \(\frac{k}{n}\).
• After removing one success, there are now \(k-1\) favourable items left out of \(n-1\) total items.
• Therefore, probability of two successes in a row is the product: \(\frac{k}{n} \times \frac{k-1}{n-1}\).

Recipe (how to use it)

1. Identify \(n\), the total items.
2. Identify \(k\), the number of favourable outcomes.
3. Compute first probability = \(k/n\).
4. Compute second probability = \((k-1)/(n-1)\).
5. Multiply them together.

Spotting it

If the problem says “without replacement”, then probabilities change after the first draw. This is the key signal.

Common pairings

• Hypergeometric distribution (general case of several successes without replacement).
• Comparisons with “with replacement” probabilities.

Mini examples

1. Example 1: Bag has 3 red, 5 blue. Probability of two reds without replacement: \(P=(3/8)\times(2/7)=6/56=3/28\).
2. Example 2: Deck of 52 cards, probability of two hearts without replacement: \(P=(13/52)\times(12/51)=156/2652=1/17\).

Pitfalls

• Forgetting that the denominator decreases on the second draw.
• Using the “with replacement” square formula by mistake.

Exam strategy

• Always check wording: “without replacement” means dependent events.
• Write both fractions separately before multiplying.
• Simplify step by step to avoid mistakes.

Summary

The formula \(P=\frac{k}{n}\times\frac{k-1}{n-1}\) calculates the probability of two dependent successes without replacement. It reflects how both numerator and denominator shrink after the first success.