Geometric Distribution Math Example 4

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Example 4

hard
A quality inspector samples items until finding the first defective. Defect probability is p=0.05p=0.05. Find P(X>10)P(X > 10) (probability of needing more than 10 inspections) and the expected number to inspect.

Solution

  1. 1
    P(X>10)=P(no defect in first 10)=(1−p)10=(0.95)10≈0.5987P(X > 10) = P(\text{no defect in first 10}) = (1-p)^{10} = (0.95)^{10} \approx 0.5987
  2. 2
    About 60% chance of needing more than 10 inspections
  3. 3
    E(X)=10.05=20E(X) = \frac{1}{0.05} = 20 items expected before first defect

Answer

P(X>10)=(0.95)10≈0.599P(X>10) = (0.95)^{10} \approx 0.599; expected inspections = 20.
With a rare defect rate (5%), it takes a long time to find the first defect. P(X > k) = (1-p)^k is the complement of finding a defect within k trials. The expected value of 20 confirms rare defects require many inspections.

About Geometric Distribution

The probability distribution for the number of independent Bernoulli trials needed to get the first success, where each trial has success probability pp.

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