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This problem involves drawing three cards from a deck of cards. Assume that the deck contains 3 aces, 5 other face cards, and 11 non-face cards, and that you randomly draw 3 cards. A random variable Z is defined to be 5 times the number of aces plus 4 times the number of other face cards drawn. How many different values are possible for the random variable Z?

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Answer with Step-by-step explanation:

We are given that

Number of ace cards=3

Number of face cards=5

Number of non-face cards=11

When we draw randomly 3 cards.

Z=5(number of aces)+4(number of other face cards)

We have to determine the possible different values for the random variable Z

Number of ace cards=3, number of other face cards=0

Z=5(3)+4(0)=15

Number of ace cards=2, number of other face cards=1

[tex]z=5(2)+4(1)=14[/tex]

Number of ace card=1, number of other face  cards=2

Z=5(1)+4(2)=13

Number of ace card=0, number of other face cards=3

Z=5(0)+4(3)=12

facundo3141592

The possible values of the random variable z are:

{0, 3, 4, 6, 7, 8, 9, 10, 11, 12}

What are the possible values of Z?

Our deck has:

  • 3 ace cards.
  • 5 other face cards.
  • 11 non face cards.

So there is a total of 19 cards.

Now we draw 3 cards, and we define the variable Z as:

Z = 3*x + 4*y

where x is the number of aces drawn, y is the number of other face cards drawn.

And the possible combinations of x and y (and the correspondent value of z) that we have are:

  • x = 3, y = 0  ⇒  z = 9
  • x = 2, y = 1   ⇒  z = 10
  • x = 1, y = 2   ⇒  z =  11
  • x = 0, y =3    ⇒  z = 12
  • x = 2, y = 0    ⇒  z = 6
  • x = 1, y = 1      ⇒  z = 7
  • x = 0, y = 2     ⇒  z = 8
  • x = 1, y = 0      ⇒  z = 3
  • x = 0, y = 1      ⇒  z = 4
  • x = 0, y = 0     ⇒  z = 0.

So the possible values of z are:

{0, 3, 4, 6, 7, 8, 9, 10, 11, 12}

If you want to learn more about random variables, you can read:

https://brainly.com/question/15246027

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