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If there are known to be 4 broken transistors in a box of 12, and 3 transistors are drawn at random, what is the probability that none of the 3 is broken?
Since there are 4 broken transistors, there must be 8 good ones. P (first pick is good) = 8/12. Of the remaining 11 transistors, 7 are good, and so P (second pick is good) = 7/12. Finally, P (third pick is good) = 6/10. Therefore, P (all three are good) = 8/12 x 7/11 x 6/10 = 14/55 = 0.255.