Question 669660
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once a batter is withdrawn it is not replaced before drawing the next one.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.