# Depleted batteries

Mathematics Asked by Francesco Totti on December 7, 2020

Accidentally, two depleted batteries got into a set of five batteries. To remove the two depleted batteries, the batteries are tested one by one in a random order. Let the random variable $$X$$ denote the number of batteries that must be tested to find the two depleted batteries. What is the probability mass function of $$X$$?

WARNING $$rightarrow$$ Do not use conditional probabilities.

If I understand correctly the text (that in my opinion is really unclear), $$X$$ represents the number of tests needed to find simultaneously the two depleted batteries. Thus, the minimum number of tests that I have to do in order to find simultaneously the two depleted batteries is two.

Let $$V$$ be the event the battery is depleted, and let $$F$$ be the event that the battery is not depleted.

$$mathbb{P}(VV)=frac{2cdot 1cdot 3!}{5!}$$ or in an alternative way $$mathbb{P}(X=2)=frac{binom{2}{2}cdot 2!cdot 1cdot 3!}{5!}$$ where:

• $$5!$$ is the number of ways to arrange in a random order all batteries on a row.

• $$binom{2}{2}$$ means that I take two depleted batteries from the group of depleted batteries. And with $$2!$$ I arranged them (ie to say $$i_1i_2$$ or $$i_2i_1$$).

• $$1$$ is the number of cases in which the couple can be located on the row of two tests ($$rightarrow$$ 1 case)

• $$3!$$ is the number of ways to arrange the other batteries.

So far, so good: the two probabilities are equal.

$$mathbb{P}[(FVV)cup (VVF)cup (VFV)]=frac{3cdot 2cdot 1cdot 2!+2cdot 1cdot 3cdot 2!+2cdot 3cdot 1cdot 2!}{5!}$$ or $$mathbb{P}(X=3)=frac{binom{2}{2}cdot 2!cdot 2cdot binom{3}{1}cdot 1!cdot 2cdot 2!+binom{2}{2}cdot 2!cdot 1cdot binom{3}{1}cdot 1!cdot 1cdot 2!}{5!}$$ where

• $$binom{2}{2}$$ means that I take two depleted batteries from the group of depleted batteries. And with $$2!$$ I arranged them (ie to say $$i_2i_3$$ or $$i_1i_2$$).

• $$2$$ is the number of cases in which the couple can be located on the row of three tests (at the right and left end of the row $$rightarrow$$ 2 cases)

• $$binom{3}{1}$$ means that I take one charged battery from the group of charged batteries. And with $$1!$$ I arranged it (ie to say $$i_1$$ or $$i_3$$).

• $$2$$ is the number of cases in which the charged battery can be located on the row of three tests (at the left and right end of the row $$rightarrow$$ 2 cases)

• $$2!$$ is the number of ways to arrange the other batteries.

• $$binom{2}{2}$$ means that I take two depleted battery from the group of depleted batteries. And with $$2!$$ I arranged them (ie to say $$i_1i_3$$ and $$i_3i_1$$).

• $$1$$ is the number of cases in which the depleted batteries can be located on the row of three tests (at the left and right end of the row $$rightarrow$$ 1 case)

• $$binom{3}{1}$$ means that I take one charged battery from the group of charged batteries. And with $$1!$$ I arranged it (ie to say $$i_2$$).

• $$1$$ is the number of cases in which the charged battery can be located on the row of three tests (in the middle of the row $$rightarrow$$ 1 cases)

• $$2!$$ is the number of ways to arrange the other batteries.

Unfortunately, in this case, the first probability is $$frac{3}{10}$$ and the second is $$frac{1}{2}$$: why?

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