$$rac{1}{2}$$
$$rac{1}{3}$$
$$rac{1}{4}$$
$$rac{1}{5}$$
Answer is Wrong!
Answer is Right!
The correct answer is $\boxed{\frac{1}{20}}$.
There are two ways to choose one defective calculator and four good calculators:
- Choose one defective calculator from the two defective calculators and then choose four good calculators from the 23 good calculators. This can be done in $\binom{2}{1}\binom{23}{4} = \frac{25 \times 24 \times 23 \times 22}{4 \times 3 \times 2 \times 1} = 2300$ ways.
- Choose four good calculators from the 23 good calculators and then choose one defective calculator from the two defective calculators. This can be done in $\binom{23}{4}\binom{2}{1} = \frac{23 \times 22 \times 21 \times 20}{4 \times 3 \times 2 \times 1} = 1380$ ways.
Therefore, the probability of choosing one defective calculator and four good calculators is $\frac{2300 + 1380}{25 \times 24} = \frac{3680}{600} = \boxed{\frac{1}{20}}$.
Option A is incorrect because it is the probability of choosing two defective
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