Question 971730: Ten light bulbs are chosen from a manufacturing lot and tested. If the probability of any bulb being defective is 0.1, find the probabilities of the following events: (a) exactly 2 of the bulbs are defective; (b) no more than 3 of the bulbs are defective; (c) at least 9 of the bulbs are not defective.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Binomial distribution:
exactly 2 bulbs defective is 10C2 (0.1)^2 (0.9)^8= 0.194.
There are 45 ways two bulbs can be defective, and that is 10C2
No more than 3 are defective.
P (0) are defective = 0.9^10=0.3486
P(1) is defective = 10 *0.9^9 *0.1 = 0.3874 (the highest, because it is expected value)
P(2) are defective=0.194 from above
P(3) are defective= 10C3*(.9)^7 * (.1)^3= 0.057
This sum is 0.987
We want NO MORE THAN this, so that is the complement. It is 0.013.
At least 9 of the bulbs are not defective. That is the probability of (1) being defective and (0) being defective, which we computed above. That is 0.7360.
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