another finite math probility problem need help?

Question

involving drawing three cards from a deck of cards. Assume that the deck contains 4 aces, 7 other face cards, and 9 non-face cards, and that you randomly draw 3 cards. A random variable Z is defined to be 3 times the number of aces plus 2 times the number of other face cards drawn.

How many different values are possible for the random variable Z?

part 1 is 9

PART 2:

Fill in the table below to complete the probability density function. Be certain to list the values of Z in ascending order

value of X probability

Answer 1

The possible values of Z are 0,1,2,3,4,5,6,7,9.

To compute the probabilities, you once again use the binomial coefficient

C(n,k) = n! / [k!(n-k)!]

that tells you the number of ways to choose k objects from a set of n.

First, compute the total number of ways to choose 3 cards from a set of 4+7+9=20. That would be

C(20,3) = 20! / [3!17!] = 18*19*20 / 3! = 1,140.

Now, compute the number of ways to obtain each possible value of Z.

Example: Z=5 if one ace is drawn, and 2 other face cards are drawn.

How many ways are there to choose one of the four aces? That would be C(4,1) = 4.

How many ways are there to choose two of the seven other face cards? That would be C(7,2) = 7!/[2!5!] = 6*7/2 = 21.

Multiply these together, and there are 84 ways to choose one ace and two other face cards.

The probability is 84/1140 = 0.0737.

All of the others can be done in a similar fashion. Keep in mind that there are two ways for Z to be 3. In that case, handle each type of scenario like the others, and then add the probabilities together

We're using these basic laws of counting number of ways to perform tasks:
1) If there are n ways to do task A, and m ways to do task B, then there are nm ways to do task A *and* task B. That is used for just about all of the possible values of Z.
2) If there are n ways to do task A, and m ways to do task B, then there are n+m ways to do task A *or* task B. That is used for the case of Z=3.

Answer 2

indexed lower than are the steps: -32=4c-12 once you turn an integer from one aspect of the equivalent signal to the different aspect, then the integer might want to substitute its signal. 12-32=4c -20=4c -20/4=c -5=c -5 is the answer