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A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws is defective will be
- a)100%
- b)50%
- c)47%
- d)49%
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A box contains 10 screws, 3 of which are defective. Two screws are dra...
A = First drawn screw non-defective.
B = Second drawn screw non-defective.
P(A) = 7/10
Because 7 out of 10 screws are non-defective and we sample at random,
P(B) = 7/10
The second drawing is the same as the beginning.
P(A Intersection B) = P(A) * P(B) = 0.7 * 0.7 = 0.49
It is Converted to percentage = 0.49 * 100 = 49%
The probability of two screws are defective is 49%.
B = Second drawn screw non-defective.
P(A) = 7/10
Because 7 out of 10 screws are non-defective and we sample at random,
P(B) = 7/10
The second drawing is the same as the beginning.
P(A Intersection B) = P(A) * P(B) = 0.7 * 0.7 = 0.49
It is Converted to percentage = 0.49 * 100 = 49%
The probability of two screws are defective is 49%.
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Community Answer
A box contains 10 screws, 3 of which are defective. Two screws are dra...
To solve this problem, we can use the concept of probability. The probability of an event occurring is defined as the number of favorable outcomes divided by the total number of possible outcomes.
The total number of screws in the box is 10, and we are drawing 2 screws at random with replacement. This means that after each screw is drawn, it is placed back in the box before the second screw is drawn.
Calculating the Probability:
1. The probability of selecting a non-defective screw on the first draw is (7 non-defective screws) / (10 total screws) = 7/10.
2. Since we are drawing with replacement, the probability of selecting a non-defective screw on the second draw is also 7/10.
Since we are interested in the probability that none of the two screws is defective, we multiply the probabilities of the individual events:
P(both screws are non-defective) = P(first screw is non-defective) * P(second screw is non-defective)
= (7/10) * (7/10)
= 49/100
= 0.49 or 49%
Therefore, the probability that none of the two screws drawn is defective is 49%.
The total number of screws in the box is 10, and we are drawing 2 screws at random with replacement. This means that after each screw is drawn, it is placed back in the box before the second screw is drawn.
Calculating the Probability:
1. The probability of selecting a non-defective screw on the first draw is (7 non-defective screws) / (10 total screws) = 7/10.
2. Since we are drawing with replacement, the probability of selecting a non-defective screw on the second draw is also 7/10.
Since we are interested in the probability that none of the two screws is defective, we multiply the probabilities of the individual events:
P(both screws are non-defective) = P(first screw is non-defective) * P(second screw is non-defective)
= (7/10) * (7/10)
= 49/100
= 0.49 or 49%
Therefore, the probability that none of the two screws drawn is defective is 49%.
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A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws is defective will bea)100% b)50% c)47%d)49% Correct answer is option 'D'. Can you explain this answer?
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