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RV-15-001 · Conditional probability from two production lines

A factory makes 20% of its components on line A and the rest on line B. On A, 4% fail inspection. On B, 6% fail inspection.

(a) A component is selected at random. Find the probability that it fails inspection.

(b) Given that the selected component fails, find the probability that it was made on line A.

(c) Estimate the number of failures in a batch of 5000 components.

Part (a) Probability Trees & Conditional Probability Booklet p. 60 & p. 62
Method
1) Multiply along branches for P(A and fail) and P(B and fail), then add
2) P(A | fail) = P(A and fail) ÷ P(fail)
3) Expected failures = N × P(fail)
Worked Example (RV-15-001 · Part (a))
A component is selected at random. Find the probability that it fails inspection.
1)
15×125\frac{1}{5}\times\frac{1}{25}
2)
45×350\frac{4}{5}\times\frac{3}{50}
3)
7125\frac{7}{125}
Ans = 7125\frac{7}{125}
Part (b) Probability Trees & Conditional Probability Booklet p. 60 & p. 62
Method
1) Multiply along branches for P(A and fail) and P(B and fail), then add
2) P(A | fail) = P(A and fail) ÷ P(fail)
3) Expected failures = N × P(fail)
Worked Example (RV-15-001 · Part (b))
Given that the selected component fails, find the probability that it was made on line A.
1)
Divide the probability of A and failure by the probability of failure.
2)
17\frac{1}{7}
Ans = 17\frac{1}{7}
Part (c) Probability Trees & Conditional Probability Booklet p. 60 & p. 62
Method
1) Multiply along branches for P(A and fail) and P(B and fail), then add
2) P(A | fail) = P(A and fail) ÷ P(fail)
3) Expected failures = N × P(fail)
Worked Example (RV-15-001 · Part (c))
Estimate the number of failures in a batch of 5000 components.
1)
280
Ans = 280