Question 1087892
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X = number of good batteries drawn


Lets define the following two events
A = event that at least one battery is bad (1 battery is bad, 2 are bad, or all three are bad)
B = event that 0 batteries are bad (all 3 batteries are good)


Note: another way to think of event A is to think of it as "2 good batteries selected, 1 good battery selected, or 0 good batteries selected".


According to the given table
<img src = "https://i.imgur.com/rwnfKoe.png">
We see that P(X = 3) = 0.216 which means P(B) = 0.216 is the probability of selecting all three batteries that are good


This means P(A) = 1-P(B) = 1-0.216 = <font color=red>0.784</font> is the final answer. It is the probability of selecting at least one bad battery (1 bad, 2 bad,or all 3 bad). 


Note how A and B are complementary events. One event or the other, but not both, must happen. This is why P(A)+P(B) = 1.


Another way to get to the answer is to add up the probabilities for X = 0 on up to X = 2. This indicates we have as little as 0 good batteries on up to as much as 2 good batteries. So we have 0.064+0.288+0.432 = <font color=red>0.784</font> leading to the same answer.

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