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.
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My Solution:

Total Pens = 12 Nos

Defective pens = 3

Remaining = 9, Therefore, selecting 2 no defective pens from 9 = 9C2 ways, and selecting 2 pens from 12 = 12C2 ways

Probability of neither pen will be defective = 9C2/12C2 = 36/66 = 6/11 Answer is C
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as already stated, we can solve it with oposite probalbility:
9/12*8/11=6/11
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If 3 out of the 12 pens are defective, 9 out of the 12 pens are not defective.

If a customer first select 1 pen at random from a box, the probability that this pen is not defective is \(\frac{9}{12}.\) If the customer then selects, at random, the second pen from the box, the probability that this pen is not defective is \(\frac{8}{11}\). Note that the customer already took 1 pen out of the box without replacement, hence, there are only 11 pens left in the box instead of 12 and the probability of selecting a pen that is not defective is 8 out of 11 and NOT 9 out of 12. (I fell for this mistake)

Therefore, the probability of taking 2 pens at random from the box that are not defective is: \(\frac{9}{12} * \frac{8}{11}= \frac{18}{33} = \frac{6}{11}.\)
Answer C.

Bunuel What if Probability (Neither pen will be defective) = 1-Probability (defective both time) = 1- (3/12)*(2/11) = 21/22 ?

I get the different yield.
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A shipment of 8 TV sets contains 2 black and white sets and 6 color sets. If 2 TV sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black and white set?

A) 1/7
B) 1/4
C) 5/14
D) 11/28
E) 13/26


Corrected the answer choices: E should read 13/28.

PROBABILITY APPROACH:

P(at leas one)=1-P(none)=1-6/8*5/7=13/28.

Bunuel I am confused with this approach.
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We can solve it using 2 methods: combinatorics and simple probability

1) Combinatorics
2C9/2C11=6/11
2) probability; 1st pen non-defective: 9/12 AND 2nd non defective 8/11
9/12*11/12=6/11

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it's without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.
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Since there are 3 defective pens from 12, the probability of selecting the first non-defective pen is 9/12 and the probability of selecting the second non-defective pen is 8/11. Thus, the probability that a customer buys 2 non-defective pens is 9/12 x 8/11 = 3/4 x 8/11 = 24/44 = 6/11.

Answer: C
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YAY !

Can anyone competent in probability questiones explain why my methos is not correct

So we have total 12 pens out of which 3 are defective and 9 are non defective

defective = D
non defective = N

ok so i pick one pen and second in succession -> 1/12 * 2/11 = 2/132 DD (i get both defective)

next attempt: 1/12 * 2/11 = 2/132 --- > ND

one more: 1/12 * 2/11 = 2/132 ----> NN

another attempt: 1/12 * 2/11 = 2/132 ---> DN

so i have 1/3 probality why ist it correct ?