SOLUTION: A box contains 1212 ​transistors, 33 of which are defective. If 33 are selected at​ random, find the probability that a. All are defective. (Type a fraction. Simplify your ans

Algebra ->  Permutations -> SOLUTION: A box contains 1212 ​transistors, 33 of which are defective. If 33 are selected at​ random, find the probability that a. All are defective. (Type a fraction. Simplify your ans      Log On


   

Question 1132411: A box contains 1212 ​transistors, 33 of which are defective. If 33 are selected at​ random, find the probability that
a. All are defective. (Type a fraction. Simplify your answer.)
b. None are defective. (Type a fraction. Simplify your answer.)
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I will assume that the problem is about 12 transistors, of which 3 are defective....

P(3 defective) = %28C%283%2C3%29%2AC%289%2C0%29%29%2FC%2812%2C3%29+=+%281%2A1%29%2F220+=+1%2F220

P(0 defective) = %28c%283%2C0%29%2AC%289%2C3%29%29%2FC%2812%2C3%29+=+%281%2A84%29%2F220+=+84%2F220+=+21%2F55

Another simple way to calculate the probabilities in this problem is to calculate the probabilities when choosing the transistors one at a time:

P(3 defective) = %283%2F12%29%2A%282%2F11%29%2A%281%2F10%29+=+6%2F1320+=+1%2F220

P(0 defective) = %289%2F12%29%2A%288%2F11%29%2A%287%2F10%29+=+504%2F1320+=+21%2F55