Examkin

Question

Difficulty: Very hardComplementary Probability and At-Least-One Scenarios

A quality control inspector reviews a shipment containing a total of NN customized components, of which exactly 3 are defective. The inspector randomly selects 3 components from the shipment one by one without replacement. If the probability that at least one of the selected components is defective is equal to 3135\frac{31}{35}, what is the value of NN?

Answer: 7

Answer

The total number of components in the shipment, NN, is 7.
To find NN, apply the complementary probability formula P(at least 1 defective)=1−P(0 defective)P(\text{at least 1 defective}) = 1 - P(\text{0 defective}). Given that P(at least 1 defective)=3135P(\text{at least 1 defective}) = \frac{31}{35}, the probability of drawing zero defective components is 1−3135=4351 - \frac{31}{35} = \frac{4}{35}. Out of NN total components, N−3N-3 are non-defective. Drawing 3 non-defective components without replacement gives P(0 defective)=(N−3)(N−4)(N−5)N(N−1)(N−2)P(\text{0 defective}) = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}. Setting this equal to 435\frac{4}{35} and testing integer values starting at N=6N=6 yields N=7N=7, since 4×3×27×6×5=24210=435\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}.

Step-by-Step Solution

1
Calculate the probability of the complementary event (selecting no defective components).
P(0 defective)=1−3135=435P(\text{0 defective}) = 1 - \frac{31}{35} = \frac{4}{35}.
Calculating 'at least one' directly requires summing three separate cases (1 defective, 2 defective, 3 defective), whereas using the complement P(at least 1)=1−P(none)P(\text{at least 1}) = 1 - P(\text{none}) requires evaluating only one scenario.
2
Formulate the algebraic expression for picking 3 non-defective components without replacement.
P(0 defective)=(N−33)(N3)=(N−3)(N−4)(N−5)N(N−1)(N−2)P(\text{0 defective}) = \frac{\binom{N-3}{3}}{\binom{N}{3}} = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}.
There are N−3N-3 non-defective components out of NN total components, and 3 are selected without replacement.
3
Equate the algebraic probability to the known complementary probability and solve for NN.
\frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)} = \frac{4}{35} \implies N = 7.
Testing N=7N = 7 yields 4×3×27×6×5=24210=435\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}. The function is strictly increasing for N≥6N \ge 6, making N=7N = 7 the unique integer solution.

Key Concept

Complementary Probability and Dependent Sampling (Without Replacement)
Rate this question

Topics

Complementary Probability and At-Least-One Scenarios

More questions from this topic