Q.45 One percent of the cars manufactured by a company are defective. What is the probability (upto
four decimals) that more than two cars are defective, if 100 cars are produced? __________
Probability More Than 2 Defective Cars in 100: Binomial Calculation (0.0794)
Discover the exact probability that more than two cars are defective when 1% are faulty in a batch of 100, using binomial distribution. This SEO-optimized guide explains the calculation step-by-step with Poisson approximation for quick insights.
Problem Overview
One percent of cars from a company are defective, so p = 0.01 for each car. For 100 independent cars (n = 100), find P(X > 2), where X is the number of defectives, up to four decimals. This follows a binomial distribution since trials are fixed, independent, with two outcomes (defective or not). The answer is 0.0794.
Binomial Probability Method
Compute P(X > 2) = 1 – [P(X=0) + P(X=1) + P(X=2)], using the formula P(X=k) = C(n,k) × pk × (1-p)n-k.
- P(X=0) = C(100,0) × 0.010 × 0.99100 ≈ 0.3660
- P(X=1) = C(100,1) × 0.011 × 0.9999 ≈ 0.3697
- P(X=2) = C(100,2) × 0.012 × 0.9998 ≈ 0.1849
Sum ≤ 2: ≈ 0.9206, so P(X > 2) = 1 – 0.9206 = 0.0794.
Poisson Approximation
With np = 1 (small p, large n), Poisson(λ=1) approximates well. P(X>2) = 1 – [e-1(1 + 1 + 0.5)] = 1 – 0.9197 ≈ 0.0803, close to exact 0.0794.
Why Binomial Fits
Conditions met: fixed n=100, constant p=0.01, independent production, binary defect outcome. Direct computation or software like SciPy confirms 0.0794.
Applications in Quality Control
Such probabilities help set quality thresholds in manufacturing, like accepting batches with ≤2 defectives (92.06% chance if true rate). For biotech fermentation yields or microbial defect rates, similar models apply.
