Two decks of cards are “matched,” that is, the order of the cards in the decks is compared by turning the cards over one by one from the two decks simultaneously; a “match” means that the two cards are identical. Show that the probability of at least one match is nearly.$\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{e}$

It has been given that two decks of cards are “matched,” that is, the order of the cards in the decks is compared by turning the cards over one by one from the two decks simultaneously; a “match” means that the two cards are identical

Probability means the chances of any event to occur is called it probability.

The Poisson distribution is given by the expression mentioned below.

$P\left(x\right)=\frac{{\mu }^x\cdot e^{-\mu }}{x!}$

Assume that there is no cards match so the average.

$\begin{array}{c}\mu =np\\ =n\left(\frac{1}{n}\right)\\ =1\end{array}$

All couples are equally likely, the probability of getting no match is given by expression mentioned below.

$\begin{array}{c}p\left(0\right)=\frac{1^0\cdot e^{-1}}{0!}\\ =e^{-1}\end{array}$

Now the probability of getting at least one couples is given by expression mentioned below.

$\begin{array}{c}p(x>0)=1-p\left(0\right)\\ =1-e^{-1}\\ =1-\frac{1}{e}\end{array}$

Hence, it has been proved.