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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 two screws is defective will be

- $100\%$
- $50\%$
- $49\%$
- None of these.

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There are $7$ non-defective balls in the box.

The probability that both the balls turn out to be non-defective would be: $\frac{^{7}C_1}{^{10}C_1}$ $\times$ $\frac{^{7}C_1}{^{10}C_1}$ $=$ $0.49$ or $49\%$

Option C is correct.

The probability that both the balls turn out to be non-defective would be: $\frac{^{7}C_1}{^{10}C_1}$ $\times$ $\frac{^{7}C_1}{^{10}C_1}$ $=$ $0.49$ or $49\%$

Option C is correct.

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P(UD) be Probability of non-defective screws=7/10

P(D) be Probability of Defective screws=3/10

D1 First defective screw drawn P(D1) = 3/10.

D2 second defective screw drawn P(D2) = 3/10.

probability that none of two screws is defective will be $P(\overline{D1}\cap \overline{D2})=P(\overline{D1})*P(\overline{D2})$ { as both events are independent }

=$\frac{7}{10} * \frac{7}{10}=\frac{49}{100}= 49%$ %

As only one two outcomes are possible so D1 or D2 complement will be non-defective screw UD1 OR UD2

P(D) be Probability of Defective screws=3/10

D1 First defective screw drawn P(D1) = 3/10.

D2 second defective screw drawn P(D2) = 3/10.

probability that none of two screws is defective will be $P(\overline{D1}\cap \overline{D2})=P(\overline{D1})*P(\overline{D2})$ { as both events are independent }

=$\frac{7}{10} * \frac{7}{10}=\frac{49}{100}= 49%$ %

As only one two outcomes are possible so D1 or D2 complement will be non-defective screw UD1 OR UD2