If 3 of 12 are defective ----> 9 of the 12 are good
Prob(2 drawn are defective) = (3/12)(2/11) = 1/22 <--- DD
Prb(2 drawn are good) = (9/12)(8/11) = 6/11<----- GG
prob(if 2 drawn, one is defective, the other is good)
= (3/12)(9/11) + (9/12)(3/11) = 9/22 <----- could be GD or DG
notice the 3 probs add up to 1, as expected.
By Binomial distribution:
1. C(3,2)/C(12,2) = 3/66 = 1/22
2. C(9,2)/C(12,2) = 36/66 = 6/11
3. C(3,1)*C(9,1)/C12,2) = 3*9/66 = 9/22 , same as above
A box contain 12 bulbs with 3 defective ones. if two bulbs are drawn from the box together,what is the probability that both bulbs are defective,both are non defective and one bulb is defective
1 answer
1 answer