Four hundred draws will be made at random with replacement from the box with the following tickets. 1 3 5 7 Estimate the chance that there will be fewer than 90 3’s. Give your answer as a percent rounded to 2 decimals.
Answer
🎲 Probability Estimate – Drawing 3’s from a Random Box
🔍 Step 1: Identify the Setup
This is a binomial probability problem where:
- n = 400 (number of trials)
- p = 1/4 = 0.25 (probability of drawing a 3 since there are 4 equally likely outcomes: 1, 3, 5, 7)
- q = 1 − p = 0.75
📐 Step 2: Use the Normal Approximation to the Binomial
Since n is large, we apply the normal approximation:
Mean (μ) = n × p = 400 × 0.25 = 100
Standard Deviation (σ) = √(n × p × q) = √(400 × 0.25 × 0.75) = √75 ≈ 8.660
Standard Deviation (σ) = √(n × p × q) = √(400 × 0.25 × 0.75) = √75 ≈ 8.660
🧮 Step 3: Apply Continuity Correction
To find the probability of getting fewer than 90 3’s, we compute:
P(X < 90) ≈ P(Y < 89.5), where Y ~ Normal(μ, σ²)
📏 Step 4: Convert to Z-score
Use the Z formula: Z = (X − μ) / σ
Z = (89.5 − 100) / 8.660 ≈ −10.5 / 8.660 ≈ −1.212
📊 Step 5: Find the Area Under the Normal Curve
Using the Z-table or calculator, we find:
P(Z < −1.212) ≈ 0.1132
✅ Final Answer:
The estimated chance is approximately 11.32%
💡 Tip: The normal approximation is a powerful tool for estimating binomial probabilities when the number of trials is large and both np and nq are at least 5.