(7/10)x(6/9) = 7/15

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A box contains 10 screws, 3 of which are defective. Two screws are drawn at random **with out replacement**. The probability that none of the two screws is defective will be ..

This is a same question from Gate 2002, but it is having a small change, can anyone help me to find method to find it's answer.

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Probability none of the screw is defective= $\frac{7}{10}*\frac{6}{9}$=$\frac{7}{15}$

Probability first one defective and second one is not defective=$\frac{3}{10}*\frac{7}{9}$=$\frac{7}{30}$

Probability first one is not defective and second one is defective=$\frac{7}{10}*\frac{3}{9}$=$\frac{7}{30}$

Probability both are defective =$\frac{3}{10}*\frac{2}{9}$=$\frac{1}{15}$