Two Successes (With Replacement)

Statement

When an experiment is repeated twice with replacement, the probability of two successes is the square of the single success probability:

\[ P = \left(\frac{k}{n}\right)^2 \]

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

Why it’s true

• The probability of success in one trial is \(\frac{k}{n}\).
• Because the trials are independent (replacement makes the total stay the same), the probability of two successes is:

\(\frac{k}{n} \times \frac{k}{n} = \left(\frac{k}{n}\right)^2\).

Recipe (how to use it)

1. Identify total possible outcomes \(n\).
2. Count favourable outcomes \(k\).
3. Find single-trial probability: \(k/n\).
4. Square it for two successes: \((k/n)^2\).

Spotting it

Look for problems mentioning two draws with replacement, or “probability both are …”. That signals squaring the single probability.

Common pairings

• Binomial probability (more than two trials).
• “Without replacement” probability (different formula: multiply adjusted fractions).

Mini examples

1. Example 1: Bag has 3 red and 5 blue balls. Find probability both chosen are red (with replacement). \(P=(3/8)^2=9/64\).
2. Example 2: A spinner has 10 equal sections, 4 marked “win”. Probability of two wins with replacement: \((4/10)^2=16/100=0.16\).

Pitfalls

• Forgetting replacement means denominator stays the same.
• Not squaring the probability.
• Mixing this with “without replacement” cases (numerators and denominators change there).

Exam strategy

• Always check wording: if it says “with replacement”, use this formula.
• Reduce fractions before squaring for easier arithmetic.
• Show working: single probability first, then square.

Summary

The formula \(P=(k/n)^2\) gives the probability of two independent successes with replacement. It’s a direct application of multiplying probabilities for independent events.