10 good and 3 bad batteries are mixed and then 5 are chosen. What is the probability of the fifth one being dead given that the first 4 aren't? I approached this as follows:
$A$: First four are good, $B$: Fifth one is bad. Then $$|A|={10\choose4}$$ because this is the number of ways we can choose 4 good batteries from the available 10. Then $$P(A)=\frac{10\choose4}{13\choose4}$$ I want to find $P(B|A)=\frac{P(A\cap B)}{P(A)}$. For this I need: $A\cap B$: First four are good and fifth one is dead. Then $|A\cap B|={10\choose4}{3\choose1}$ because this is the number of ways we can choose 4 good batteries from the available 10 followed by choosing 1 of the available 3 dead ones. Then $$P(A\cap B)=\frac{{10\choose4}\cdot{3\choose1}}{13\choose5}$$ However, once a substitute these values into the conitional probability expression I get $\frac53$.
 A: $\binom{13}{5}$ counts the number of ways of selecting five batteries where we do not care in which order they are selected.  All ways of selecting the same five batteries are considered "the same" regardless of if it was a bad one first followed by four good, or a bad one last, etc...  Recognize that "First four are good and Fifth one is bad" makes explicit reference to the order in which the batteries are drawn, so your sample space must also make explicit reference to the order in which the batteries are drawn to be consistent.
$\Pr(A\cap B) = \dfrac{\binom{10}{4}\cdot\binom{3}{1}}{\binom{13}{4}\binom{9}{1}}$ where here the denominator is in reference to "draw four batteries and then draw a fifth."   Equivalently, $\Pr(A\cap B)=\dfrac{\binom{10}{4}\cdot\binom{3}{1}}{\binom{13}{5}\cdot 5}$ where the denominator is in reference to "draw five batteries and then pick which of the five was to be the 'fifth' battery."  That is to say, your answer was off by a factor of five.  Correcting this, we get an answer of $\frac{5}{3}\cdot\frac{1}{5}=\frac{1}{3}$
We could just as well have kept track of the exact order in which every battery was drawn, leading to $\Pr(A\cap B) = \dfrac{10\cdot 9\cdot 8\cdot 7\cdot 3}{13\cdot 12\cdot 11\cdot 10\cdot 9}$ which eventually yields the same answer.  The punchline, again, is that however detailed you are with your numerator in terms of keeping track of the order of things you must be just as detailed with the denominator in the same way.
Even simpler than doing all of this however is to just find the conditional probability directly through intuitive reasoning.  The probability the fifth is a bad battery given the first four aren't... well, that is just the same as asking the probability of having drawn a bad battery out of a bag with 6 good and 3 bad batteries in it in the first place.  That is, we simply imagine we are in the scenario where the event we are conditioning has already occurred and we look at what probabilities are like given the result.  The answer is that with 6 good and 3 bad batteries and us pulling a single battery asking if it is bad... the probability will very simply be:
$$\frac{3}{9}=\frac{1}{3}$$
same as the corrected answer above.
A: Hint: The solution is much, much, much easier. One division is all.
After choosing four working batteries, you have 3 bad batteries and 6 good batteries left, 9 batteries in total. The probability that the next one is bad is 3 / 9. The probability that the next one is good is 6 / 9.
