Question:
A box contains 20 thermometers, 3 of which are defective. One person randomly draws 2 thermometers from the box, one by one, without replacement. The probability in percent (rounded off to two decimal places) that none of these TWO thermometers is defective, is ______ %.
A box contains 20 thermometers, 3 of which are defective. One person randomly draws 2 thermometers from the box, one by one, without replacement. The probability in percent (rounded off to two decimal places) that none of these TWO thermometers is defective, is ______ %.
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Use the multiplication rule for dependent events: P(1st good) times P(2nd good given 1st good), then convert to a percentage.
Updated On: Jul 28, 2026
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Correct Answer: 71
Solution and Explanation
Step 1: Set up the problem.
There are 20 thermometers in total, 3 defective and 17 good (non defective). Two are drawn one by one without replacement. We need the probability that both draws give a good thermometer.
Step 2: Find the probability the first thermometer is good.
Out of 20 thermometers, 17 are good, so
\[ P(\text{1st good})=\frac{17}{20} \]
Step 3: Find the probability the second thermometer is also good, given the first was good.
After removing one good thermometer, 19 thermometers remain, of which 16 are good. So
\[ P(\text{2nd good}\mid \text{1st good})=\frac{16}{19} \]
Step 4: Multiply the two probabilities.
Since the draws are without replacement, the events are dependent, so use the multiplication rule:
\[ P(\text{both good})=\frac{17}{20}\times\frac{16}{19}=\frac{272}{380} \]
Step 5: Simplify and convert to a percentage.
\[ \frac{272}{380}=\frac{68}{95}\approx0.715789 \]
\[ 0.715789\times100\approx71.58\% \]
Step 6: Final answer.
Rounded to two decimal places, the probability is
\[ \boxed{71.58\%} \]
There are 20 thermometers in total, 3 defective and 17 good (non defective). Two are drawn one by one without replacement. We need the probability that both draws give a good thermometer.
Step 2: Find the probability the first thermometer is good.
Out of 20 thermometers, 17 are good, so
\[ P(\text{1st good})=\frac{17}{20} \]
Step 3: Find the probability the second thermometer is also good, given the first was good.
After removing one good thermometer, 19 thermometers remain, of which 16 are good. So
\[ P(\text{2nd good}\mid \text{1st good})=\frac{16}{19} \]
Step 4: Multiply the two probabilities.
Since the draws are without replacement, the events are dependent, so use the multiplication rule:
\[ P(\text{both good})=\frac{17}{20}\times\frac{16}{19}=\frac{272}{380} \]
Step 5: Simplify and convert to a percentage.
\[ \frac{272}{380}=\frac{68}{95}\approx0.715789 \]
\[ 0.715789\times100\approx71.58\% \]
Step 6: Final answer.
Rounded to two decimal places, the probability is
\[ \boxed{71.58\%} \]
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