A standard deck of cards is​ purchased, and the deck includes 3 jokers. This means if the jokers are​ included, the deck contains 5 5 cards. Teresa is playing a game in which she is dealt 8 cards from a deck that includes the jokers. What is the probability that​ Teresa's hand includes at least one​ joker?

Answer:

The probability is [tex]0.3819[/tex]

Step-by-step explanation:

We know that Teresa is playing a game in which she is dealt 8 cards from a deck that includes 3 jokers and 52 common cards (a total of 55 cards).

Let's define the following random variable :

[tex]X[/tex] : '' Number of jokers in the hand of Teresa ''

We need to find [tex]P(X\geq 1)[/tex]

This probability is equivalent to :

[tex]P(X\geq 1)=1-P(X=0)[/tex]

[tex]P(X=0)[/tex] is the probability of having none jokers in the 8 card hand.

In order to find [tex]P(X=0)[/tex] we are going to count all the cases in which [tex]X=0[/tex] (given that we are in presence of an equally - likely sample space)

We calculate [tex]P(X=0)[/tex] as :

[tex]P(X=0)=\frac{\left(\begin{array}{c}52&8\end{array}\right)\left(\begin{array}{c}3&0\end{array}\right)}{\left(\begin{array}{c}55&8\end{array}\right)}[/tex]

We define the combinatorial number [tex]nCr=\left(\begin{array}{c}n&r\end{array}\right)=\frac{n!}{r!(n-r)!}[/tex]

In the denominator we have [tex]\left(\begin{array}{c}55&8\end{array}\right)[/tex] which represents all the ways in which we can extract 8 cards from the deck of 55 cards.

In the numerator we have the product of [tex]\left(\begin{array}{c}52&8\end{array}\right)[/tex]  (which represents all the ways in which we can choose 8 cards from the 52 common cards) and   [tex]\left(\begin{array}{c}3&0\end{array}\right)[/tex]  (which represents that from the total of 3 jokers we extract 0)

 If we perform the operation we find that :

[tex]P(X=0)=0.6181[/tex]

Finally,

[tex]P(X\geq 1)=1-P(X=0)=1-0.6181=0.3819[/tex]

The probability is [tex]0.3819[/tex]

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