SOLUTION: you choose 3 batteries at random (with replacement) from a bag containing 10 batteries. Let x = the number of good batteries drawn. x/p(x) 0/.064 1/.288 2/.432 3/.216 What is the

Click here to see ALL problems on Probability-and-statistics

Question 1087892: you choose 3 batteries at random (with replacement) from a bag containing 10 batteries. Let x = the number of good batteries drawn. x/p(x) 0/.064 1/.288 2/.432 3/.216 What is the probability that at least one battery drawn is bad?

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!

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

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 = 0.784 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 = 0.784 leading to the same answer.