SOLUTION: if there are known to be 4 broken transistors in a box of 12 and 3 transistor are drawn at random what is probability that none of the three is broken? a)0.375 b)0.255 c)2.55

Question 697651: if there are known to be 4 broken transistors in a box of 12 and 3 transistor are drawn at random what is probability that none of the three is broken?
a)0.375
b)0.255
c)2.55
d)0.75
e)0.556
Answer by Positive_EV(69) (Show Source): You can put this solution on YOUR website!
The probability that there are no broken transistors is the probability that the first transistor is not broken times the probability that the second transistor is not broken times the probability that the third transistor is not broken. Since each transistor is removed from the box, the transistors are drawn without replacement.

The probability that the first transistor is not broken is equal to the number of working transistors times the total number of transistors in the box, which is 8/12 = 2/3.

Since one working transistor is removed from the box, there are now 11 transistors in the box, of which 7 work. The probability that the second transistor works is thus 7/11.

This transistor is also removed, which means there are now 6 working transistors left in the box, and 10 total transistors. The probability that the third transistor works is 6/10 = 3/5.

The probability that all three transistors work is the product of these probabilities, which is (2/3)*(7/11)*(3/5) = 42/165 = 14/55 = .255.
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