6,937 views
5 votes
5 votes
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

3 Answers

5 votes
5 votes

http://www.emathzone.com/tutorials/basic-statistics/sampling-with-replacement.html

Probability for the first screw being non-defective, P(A)=n(e)/n(s) = 7/10

Probability for the second screw being non-defective, P(B)=n(e)/n(s) = 7/10

Answer = P(A intersection B)=P(A)*P(B) (as events are independent) = 49/100

0 votes
0 votes
if defect will be faliure then P(F)=3/10. Not defect(success) P(S)=7/10..

P(None is defect)=P(1st drawing not defect)*P(2nd drawing not defect)=P(S)*P(S)=7/10*7/10=49/100.
0 votes
0 votes
Without replacement:

$\frac{7}{10}*\frac{6}{9}=\frac{42}{90}$

With replacement:

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

After reading the question, one first assumes that the answer must be the one without replacement. But there was no option of 46.66% when it was asked in 2003. So from the options we conclude that it was asked with replacement of screws in mind.

Related questions

2 votes
2 votes
1 answer
2
practicalmetal asked Jan 3
189 views
8 pairs of hand gloves are on a shelf ( each of different colour). Four gloves are selected at random, the probability that there will be at least one pair is?