SOLUTION: There are two defective batteries in one package of six batteries and three defective batteries in the second package of six. If two batteries are selected at random from each pac

Question 669660: There are two defective batteries in one package of six batteries and three defective batteries in the second package of six. If two batteries are selected at random from each package, find the probability that at least three of the four chosen batteries will work.
I know what this is asking for I just can't seem to figure out how to set it up. Thanks for any help I may receive!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
this is what i think it might be.
you have 2 packages.
package 1 has 2 defective batteries.
package 2 has 3 defective batteries.
this is a no replacement situation.
once a batter is withdrawn it is not replaced before drawing the next one.

for package 1:
the probability of getting 0 good batteries is 2/6 * 1/5 = 2/30
the probability of getting 1 good battery is 4/6 * 2/5 * 2 = 16/30
the probability of getting 2 good batteries is 4/6 * 3/5 = 12/30
total probability is 1 as it should be.

for package 2:
the probability of getting 0 good batteries is 3/6 * 2/5 = 6/30
the probability of getting 1 good battery is 3/6 * 3/5 * 2 = 18/30
the probability of getting 2 good batteries is 3/6 * 2/5 = 6/30
the total probability is 1 as it should be.

now you want to know the probability of getting at least 3 good batteries.
this can happen in the following ways.

2 good batteries from package 1 and 1 good battery from package 2
2 good batteries from package 1 and 2 good batteries from package 2
1 good battery from package 1 and 2 good batteries from package 2.

those are the only ways it can happen.

the probability of getting 2 good batteries from package 1 and 1 good battery from package 2 is equal to 12/30 * 18/30 = 216 / 900
we'll call this probability (A)

the probability of getting 2 good batteries from package 1 and 2 good batteries from package 2 is equal to 12/30 * 6/30 = 73/900
we'll call this probability (B)

the probability of getting 1 good battery from package 1 and 2 good batteries from package 2 is equal to 16/30 * 6/30 = 96/900
we'll call this probability (C)

the total probability of getting at least 3 good batteries is equal to p(A) or p(B) or p(C) which is equal to p(A) + p(B) + p(C) since these are mutually exclusive events.

Mutually exclusive events means that A and B or A and C or B and C or A and B and C cannot exist at the same time.
In other words you can't get 2 good batteries from package 1 and 1 good battery from package 2 while at the same time get 1 good battery from package 1 and 2 good batteries from package 2. you can get one or the other but you can't get both at the same time.

the total probability, if i am guessing correctly, should be:

p(A) + p(B) + p(C) = 216/900 + 73/900 + 96/900 which is equal to 385/900.

the probability of getting at least 3 good batteries is 385/900 which is equivalent to .428 which you can round to .43.

i think this is the right way to look at it, but make sure you agree before going with it.

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