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
for the first addend
$binom{2}{2}$ 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.
and for the second addend
$binom{2}{2}$ 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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