A box contains 10 screws, 3 of which are defective. Two screws are dra...
We know that for these type of questions we solve with the help of combinations.
As for a non defective screws,they could be selected as 7C2 out of the total of 10 screws(10C2). So dividing 7C2 by 10C2, we get 0.47 or 47%
A box contains 10 screws, 3 of which are defective. Two screws are dra...
Probability of selecting a non-defective screw at random from the box = 7/10 = 0.7
Probability of selecting a defective screw at random from the box = 3/10 = 0.3
Since we are drawing two screws at random with replacement, the probability of not selecting a defective screw in the first draw is 0.7, and the probability of not selecting a defective screw in the second draw is also 0.7.
To find the probability that none of the two screws is defective, we multiply the probabilities of not selecting a defective screw in both draws.
Probability that none of the two screws is defective = 0.7 * 0.7 = 0.49 = 49%
Therefore, the correct answer is option D, 47%.
Explanation:
- The problem involves drawing two screws at random with replacement from a box of 10 screws, 3 of which are defective.
- To find the probability that none of the two screws is defective, we need to calculate the probability of not selecting a defective screw in both draws.
- The probability of not selecting a defective screw in the first draw is 7/10 since there are 7 non-defective screws out of 10 in the box.
- Since the screws are drawn with replacement, the probability of not selecting a defective screw in the second draw is also 7/10.
- To find the probability of both events happening, we multiply the probabilities of not selecting a defective screw in both draws, which gives us 0.7 * 0.7 = 0.49 = 49%.
- Therefore, the probability that none of the two screws is defective is 49%.
- The correct answer is option D, 47%.
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