A quality control inspector is reviewing a batch of 12 electronic components, of which 8 a… — Armed Services Vocational Aptitude Battery (ASVAB)

A quality control inspector is reviewing a batch of 12 electronic components, of which 8 are functioning properly and 4 are defective. If the inspector randomly selects 3 components from the batch one at a time without replacement, what is the probability that exactly 2 of the selected components are functioning properly and 1 is defective?

$\frac{28}{55}$ Answer
B
$\frac{28}{165}$
C
$\frac{4}{9}$
D
$\frac{19}{33}$

Answer

The correct probability is

\frac{28}{55}

The correct answer accounts for both the dependent nature of the events (drawing without replacement) and the multiple combinations that satisfy the condition (3 different orders).

Step-by-Step Solution

Determine the probabilities of drawing 2 functioning (F) and 1 defective (D) component in one specific order, such as F, then F, then D.

P(F_1) = \frac{8}{12}

P(F_2) = \frac{7}{11}

, and

P(D_3) = \frac{4}{10}

Because the components are drawn without replacement, the total pool (denominator) and the pool of available specific components (numerator) decrease with each successful draw.

Calculate the joint probability of this specific sequence by multiplying the individual probabilities.

\frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} = \frac{2}{3} \times \frac{7}{11} \times \frac{2}{5} = \frac{28}{165}

The probability of a sequence of dependent events occurring together is the product of their conditional probabilities.

Identify how many different sequences can result in exactly 2 functioning and 1 defective component.

There are 3 possible sequences: F-F-D, F-D-F, and D-F-F.

The single defective component can be the first, second, or third one drawn.

Calculate the total probability of achieving this outcome by multiplying the probability of a single sequence by the number of possible sequences.

3 \times \frac{28}{165} = \frac{84}{165}

Each of the 3 valid sequences is mutually exclusive and has the exact same probability of occurring, so we can sum their probabilities (or multiply by 3).

Simplify the final fraction.

\frac{84}{165}

simplifies to

\frac{28}{55}

by dividing the numerator and denominator by 3.

Fractions should always be presented in their simplest form.

Key Concept

Compound Probability of Dependent Events with Permutations

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