B. Do as Indicated! 1. Three balls are drawn in succession without replacement from a box containing 5 red balls, 6 yellow balls and 6 blue balls. Let $$ Z $$ be the random variable representing the number of blue balls. Find the values of the random variable $$ Z $$. Construct probability distribution and histogram.

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1. **Determine the Total Number of Ways to Draw 3 Balls**

The box contains 5 red, 6 yellow, and 6 blue balls. Thus, the total number of balls is
   $$
   5 + 6 + 6 = 17.
   $$
   The total number of ways to draw 3 balls (without replacement) is given by
   $$
   \binom{17}{3} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680.
   $$

2. **Possible Values of the Random Variable $$ Z $$**

The random variable $$ Z $$ represents the number of blue balls drawn. Since 3 balls are drawn, the possible values are
   $$
   Z = 0,\ 1,\ 2,\ 3.
   $$

3. **Calculate the Probability Distribution**

- **For $$ Z = 0 $$ (No Blue Balls Drawn):**

There are $$ 5 + 6 = 11 $$ non-blue balls (red and yellow). The number of ways to choose 3 non-blue balls is
     $$
     \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165.
     $$
     Thus, the probability is
     $$
     P(Z=0) = \frac{165}{680}.
     $$

- **For $$ Z = 1 $$ (Exactly 1 Blue Ball Drawn):**

Choose 1 blue ball from 6 and 2 non-blue balls from 11. The number of ways is
     $$
     \binom{6}{1} \times \binom{11}{2} = 6 \times \frac{11 \times 10}{2 \times 1} = 6 \times 55 = 330.
     $$
     Thus,
     $$
     P(Z=1) = \frac{330}{680}.
     $$

- **For $$ Z = 2 $$ (Exactly 2 Blue Balls Drawn):**

Choose 2 blue balls from 6 and 1 non-blue ball from 11. The number of ways is
     $$
     \binom{6}{2} \times \binom{11}{1} = 15 \times 11 = 165.
     $$
     Thus,
     $$
     P(Z=2) = \frac{165}{680}.
     $$

- **For $$ Z = 3 $$ (All Blue Balls Drawn):**

Choose 3 blue balls from 6. The number of ways is
     $$
     \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.
     $$
     Thus,
     $$
     P(Z=3) = \frac{20}{680}.
     $$

4. **Simplify the Probabilities**

- $$ P(Z=0) = \dfrac{165}{680} = \dfrac{33}{136} $$
   - $$ P(Z=1) = \dfrac{330}{680} = \dfrac{33}{68} $$
   - $$ P(Z=2) = \dfrac{165}{680} = \dfrac{33}{136} $$
   - $$ P(Z=3) = \dfrac{20}{680} = \dfrac{1}{34} $$

The probability distribution for $$ Z $$ is summarized in the following table:

| $$ Z $$ | Number of Ways                      | Probability $$ P(Z=z) $$      |
   |:-------:|:------------------------------------:|:-----------------------------:|
   | 0       | $$ \binom{11}{3} = 165 $$            | $$ \frac{33}{136} $$          |
   | 1       | $$ \binom{6}{1}\binom{11}{2} = 330 $$ | $$ \frac{33}{68} $$           |
   | 2       | $$ \binom{6}{2}\binom{11}{1} = 165 $$ | $$ \frac{33}{136} $$          |
   | 3       | $$ \binom{6}{3} = 20 $$              | $$ \frac{1}{34} $$            |

5. **Construct the Histogram**

To construct the histogram, plot the values of $$ Z $$ on the horizontal axis (0, 1, 2, 3) and their corresponding probabilities on the vertical axis.

- At $$ Z = 0 $$, the height is $$ \frac{33}{136} $$.
   - At $$ Z = 1 $$, the height is $$ \frac{33}{68} $$.
   - At $$ Z = 2 $$, the height is $$ \frac{33}{136} $$.
   - At $$ Z = 3 $$, the height is $$ \frac{1}{34} $$.

The histogram will display bars corresponding to these probabilities for $$ Z = 0, 1, 2, $$ and $$ 3 $$.

Thus, the probability distribution and histogram for $$ Z $$ are complete.
The random variable $$ Z $$ can take the values 0, 1, 2, or 3. The probability distribution is as follows: - $$ P(Z=0) = \frac{33}{136} $$ - $$ P(Z=1) = \frac{33}{68} $$ - $$ P(Z=2) = \frac{33}{136} $$ - $$ P(Z=3) = \frac{1}{34} $$ A histogram can be constructed with these probabilities for each value of $$ Z $$.

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