Math Problem Statement

A company receives a very large shipment of components. A random sample of 20 of these components will be checked, and the shipment will be accepted if fewer than 2 of these components are defective. What is the probability of accepting a shipment containing 5% defectives?

Solution

This problem can be modeled using the binomial distribution because we have a fixed number of trials (20 components), two possible outcomes for each trial (defective or not defective), and a constant probability of a defective component (5% or 0.05).

Let XX be the random variable representing the number of defective components in the sample of 20 components. XX follows a binomial distribution: XBinomial(n=20,p=0.05)X \sim \text{Binomial}(n = 20, p = 0.05) Where:

  • n=20n = 20 (sample size),
  • p=0.05p = 0.05 (probability of a defective component).

The shipment is accepted if fewer than 2 components are defective, meaning we want to find the probability that X<2X < 2.

This can be expressed as: P(X<2)=P(X=0)+P(X=1)P(X < 2) = P(X = 0) + P(X = 1)

Using the binomial probability formula: P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k} Where:

  • (nk)\binom{n}{k} is the binomial coefficient,
  • pkp^k is the probability of having exactly kk defective components,
  • (1p)nk(1 - p)^{n - k} is the probability of the remaining nkn - k components being non-defective.

Let's calculate P(X=0)P(X = 0) and P(X=1)P(X = 1).

  1. For X=0X = 0: P(X=0)=(200)(0.05)0(0.95)20=1×1×(0.95)20P(X = 0) = \binom{20}{0} (0.05)^0 (0.95)^{20} = 1 \times 1 \times (0.95)^{20}

  2. For X=1X = 1: P(X=1)=(201)(0.05)1(0.95)19=20×0.05×(0.95)19P(X = 1) = \binom{20}{1} (0.05)^1 (0.95)^{19} = 20 \times 0.05 \times (0.95)^{19}

Let's now compute the final probability.The probability of accepting a shipment containing 5% defectives, by finding fewer than 2 defective components in the random sample of 20, is approximately 0.736 or 73.6%.

Would you like further details or have any other questions? Here are five related questions you might find helpful:

  1. How would the probability change if the sample size increased to 30 components?
  2. What happens to the probability if the defective rate increases to 10%?
  3. How does the binomial distribution change if we allow up to 3 defective components for acceptance?
  4. Can this problem be solved using a Poisson approximation instead of a binomial distribution?
  5. How would the company optimize the sample size to minimize the risk of accepting a bad shipment?

Tip: When working with small probabilities and large sample sizes, consider using approximations like the Poisson distribution for efficiency.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
P(X < 2) = P(X = 0) + P(X = 1)

Theorems

Binomial Theorem

Suitable Grade Level

College/Advanced High School