# Probability/combinatorics word problem

An absent-minded professor goes for a walk carrying a digital audio device using 2 batteries. He has 2 fresh replacement batteries stashed away in one of four pockets. Sure enough, both batteries lose their charge and he removes them. Not wanting to throw the depleted batteries into the woods, he places them into a pocket chosen at random from the 4 available. A little while later he remembers the two fresh batteries, but he cannot remember which pocket. He fishes around in his pockets until he finds one with batteries (either 2 or 4 indistinguishable batteries). He removes 2 batteries and inserts them into the digital audio device. The digital audio device requires at least one good battery in order to play. Find the probability the digital audio device works on the first try.

I believe the denominator should be $(4)_2$, but I am having trouble with what the numerator should be.

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you can choose the batteries so that both of them are dead batteries, and there is only 1 choice for it. so the numerator is $(4)_2$ -1 – Long Mai Mar 11 '12 at 11:56
Is there a typo in this problem in that the professor has 4 good batteries in one pocket at the start (not 2 as the problem statement says), and then after a short while and the first replacement and the second failure, we have a situation of either 2 good batteries in one pocket and 2 bad in another or 4 batteries in one pocket of which 2 are good and 2 bad? Else the statement He has 2 fresh replacement batteries stashed away in one of four pockets. would imply that after the second failure, he has 4 dead batteries on him. – Dilip Sarwate Mar 11 '12 at 13:07

I don’t offhand see any way to avoid splitting the calculation into two cases depending on whether the four batteries are in a single pocket or not.

The probability that the four batteries share a pocket is $\frac14$. From $4$ batteries one can draw any of $\binom42=6$ pairs, only one of which includes no working battery, so the probability of getting a working pair is $\frac56$.

The probability that the dead batteries are not in the same pocket as the good ones is $\frac34$. What’s the probability that he picks the pocket with the good batteries first?

Just finish the two cases and combine to get the desired probability.

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There is often no single "formula" that will solve a problem. Breaking things up into natural cases can be the best way, or even the only way, to solve a problem.

For your battery problem, one can frame the analysis given by Brian M. Scott slightly differently. Let $W$ be the event "the digital device works $\dots$". We want to find the probability of $W$. It is easier to calculate the probability of the complement of $W$, the probability the device doesn't work.

The device doesn't work if (i) the professor placed the dead batteries in the "wrong" pocket and (ii) when he removed batteries from that pocket one at a time, he got a dead battery, and then another dead battery.

The probability the dead batteries are placed in the wrong pocket is $\dfrac{1}{4}$. Given that the dead batteries are placed in the wrong pocket, the probability that the first battery chosen is dead is $\dfrac{2}{4}$. And given that this happened, the probability that the second battery chosen is dead is $\dfrac{1}{3}$. So the probability that the device does not work is $$\frac{1}{4}\cdot \frac{2}{4}\cdot \frac{1}{3},$$ and therefore the probability that the device works is $$1- \left(\frac{1}{4}\cdot \frac{2}{4}\cdot \frac{1}{3}\right).$$

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The correct answer to this problem is 7/12. I say this authoritatively as the absent-minded prof who wrote the problem originally. The solution is most easily arrived at using a tree diagram. The two branches that provide a working device are:

a) good and bad batteries in separate pockets. This has probability 3/4. The conditional probability of choosing the right pocket in this case is 1/2. This case has prob 3/4x1/2 = 3/8.

b) All four in same pocket. This has prob 1/4. Number of outcomes is 4C2 = 6. The # of outcomes with at least one good battery is 5/6. This case has prob 1/4x5/6 = 5/24.

The total probability of at least one good battery is the sum, 3/8+5/24 = 14/24 = 7/12.

See Asimow/Maxwell Probability and Statistics (ACTEX 2010) Chapter 2 and accompanying Solution Manual for many more probability problems of this type.

P.S. The prof has long since switched to a device with rechargeable batteries - another solution to the problem!

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P.P.S. If only he could remember where he put the charger. – paw88789 Oct 31 '14 at 15:35