A quality control inspector evaluates components from two separate production lines. Line … — GMAT

A quality control inspector evaluates components from two separate production lines. Line 1 produces a batch of $12$ components containing $4$ defective and $8$ non-defective components. Line 2 produces a batch of $10$ components containing $2$ defective and $8$ non-defective components. The inspector randomly selects $2$ components from Line 1's batch without replacement, and then randomly selects $1$ component from Line 2's batch. What is the probability that exactly two of the three selected components are defective?

$\frac{28}{165}$ Cevap
B
$\frac{8}{45}$
C
$\frac{4}{33}$
D
$\frac{4}{55}$
E
$\frac{12}{77}$

Cevap

The probability that exactly two of the three selected components are defective is

\frac{28}{165}

To find the probability of selecting exactly two defective components across both production lines, we sum the probabilities of two mutually exclusive events: (1) selecting two defective components from Line 1 (without replacement) and one non-defective component from Line 2, which has probability

\frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} = \frac{12}{165}

; and (2) selecting one defective and one non-defective component from Line 1 and one defective component from Line 2, which has probability

2 \times \frac{4}{12} \times \frac{8}{11} \times \frac{2}{10} = \frac{16}{165}

. Adding these gives

\frac{12}{165} + \frac{16}{165} = \frac{28}{165}

Adım Adım Çözüm

Identify the mutually exclusive cases that yield exactly two defective components.

Case 1: Draw 2 defective components from Line 1 AND 1 non-defective component from Line 2.
Case 2: Draw 1 defective and 1 non-defective component from Line 1 AND 1 defective component from Line 2.

The overall target outcome can be satisfied in two distinct ways depending on which line provides the defective components.

Calculate the probability for Case 1.

P(\text{Case 1}) = \left(\frac{4}{12} \times \frac{3}{11}\right) \times \frac{8}{10} = \frac{1}{11} \times \frac{4}{5} = \frac{4}{55} = \frac{12}{165}

Drawing without replacement from Line 1 gives a probability of

\frac{4}{12} \times \frac{3}{11} = \frac{1}{11}

. Multiplying by the probability of picking a non-defective component from Line 2 (

\frac{8}{10}

) yields

\frac{4}{55}

Calculate the probability for Case 2.

P(\text{Case 2}) = \left(2 \times \frac{4}{12} \times \frac{8}{11}\right) \times \frac{2}{10} = \frac{16}{33} \times \frac{1}{5} = \frac{16}{165}

The probability of getting 1 defective and 1 non-defective component from Line 1 in 2 draws without replacement is

2 \times \frac{4}{12} \times \frac{8}{11} = \frac{16}{33}

. Multiplying by the probability of picking a defective component from Line 2 (

\frac{2}{10}

) yields

\frac{16}{165}

Sum the probabilities of Case 1 and Case 2.

P(\text{Total}) = \frac{12}{165} + \frac{16}{165} = \frac{28}{165}

Since the two cases are mutually exclusive, their individual probabilities are added.

Anahtar Kavram

Probability of Independent and Dependent Events

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