In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?
A. 1/6
B. 2/9
C. 6/11
D. 9/16
E. 3/4
Kudos for a correct solution.
Bunuel wrote:
In a box of 12 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?
A. 1/6
B. 2/9
C. 6/11
D. 9/16
E. 3/4
Kudos for a correct solution.
Method- 1There are 9 fine pieces of pen and 3 defective in a lot of 12 pens
i.e. Probability of first pen NOT being defective = (9/12)
i.e. Probability of Second pen NOT being defective = (8/11) [11 pen remaining with 8 defective remaining considering that first was defective]
Probability of Both pen being NON-defective = (9/12)*(8/11) = 6/11
Answer: option C
Method- 2There are 9 fine pieces of pen and 3 defective in a lot of 12 pens
No. of ways of choosing 2 NON defective out of 9 fine pieces of pen = 9C2
No. of ways of choosing 2 out of 12 pieces of pen = 12C2
Required probability that none of the chosen is defective = 9C2 / 12C2 = 36 / 66 = 6/11
Answer: option C