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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

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