A package contains 10 resistors, 2 of which are defective. If 3 are selected, find the probability of getting 1 defective resistor.
A) $\approx 0.3571$
B) $\approx 0.4667$
C) $\approx 0.0167$
D) $\approx 0.2333$

Question

A package contains 10 resistors, 2 of which are defective. If 3 are selected, find the probability of getting 1 defective resistor. 
A) $\approx 0.3571$
B) $\approx 0.4667$
C) $\approx 0.0167$
D) $\approx 0.2333$

Quizplus · Accepted Answer

The probability of selecting 1 defective resistor out of 3 chosen from 10, where 2 are defective, can be calculated using combinations. The number of ways to choose 1 defective out of 2 is $C(2,1)$, and the number of ways to choose 2 non-defective out of 8 is $C(8,2)$. The total number of ways to choose any 3 resistors out of 10 is $C(10,3)$. Thus, the probability is $\frac{C(2,1) \times C(8,2)}{C(10,3)} = \frac{2 \times 28}{120} = \frac{56}{120} = 0.4667$.

A Package Contains 10 Resistors, 2 of Which Are Defective ≈0.3571\approx 0.3571≈0.3571

A Package Contains 10 Resistors, 2 of Which Are Defective $\approx 0.3571$