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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

  1. $100\%$
  2. $50\%$
  3. $49\%$
  4. 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.
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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
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C. 49% ...

 

Total number of screws =10 ...

Defective screws =3 ...

We have to find the probability that none of them is defective ...

Number of ways of selecting 2 screws from 10 screws with replacement

=(10C1)× (10C1).…

 

Hence, The probability that none of them is defective, if the sample is drawn with replacement

=(7C1)/(10C1) . (7C1)/ (10C1)

=(7/10) . (7/10)

=49/100 .…

 

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