Practice 2 Solutions

State whether the scenario describes an independent or dependent event.

A teacher is assigning the order in which students will present their projects this month. Once a student presents, their turn is complete for that month.

Dependent

Three experiments were conducted simultaneously: flipping a coin, rolling one die, and using a spinner.

Independent

A baseball team calculated its batting average after calculating the individual batting averages of its top nine players.

Dependent

For problems 4–7, use the spinner and the eight-sided die with numbers 1 through 8 to find the probability.

Rolling an even number and landing on yellow

$\begin{array}{rcl} P (even, Y) & = & \frac{4}{8} \cdot \frac{1}{5} \end{array}$

$\frac{1}{10}$

Note

Remember that in this problem the die has eight sides.

Rolling a multiple of 3 and landing on a primary color

$\begin{array}{rcl} P (mult 3, primary) & = & \frac{2}{8} \cdot \frac{3}{5} \end{array}$

$\frac{3}{20}$

Note

Primary colors are red, blue, and yellow.

Multiples of 3 are 1 and 3.

Rolling a 5 and not landing on purple

$P (5, not purple) = \frac{1}{8} \cdot \frac{4}{5}$

$\frac{1}{10}$

Rolling any number except 7 and landing on purple or green

$\begin{array}{rcl} P (not 7, purple or green) & = & \frac{7}{8} \cdot \frac{2}{5} \end{array}$

$\frac{7}{20}$

For problems 8–11, use the following scenario.

Cards are drawn from a standard deck with replacement. Determine the percent chance to the nearest hundredth of a percent.

Heart, then a five

$\begin{array}{rcl} P (♥ \cap 5) & = & \frac{13}{52} \cdot \frac{4}{52} \\ = & \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52} \\ = & 0.0192 \end{array}$

1.92%

$P (10, A)$

$\begin{array}{rcl} P (10, A) & = & \frac{4}{52} \cdot \frac{4}{52} \\ = & \frac{1}{13} \cdot \frac{1}{13} \\ = & \frac{1}{169} \\ = & 0.0059 \end{array}$

0.59%

Three face cards in a row

$\begin{array}{rcl} P (face, face, face) & = & {(\frac{12}{52})}^{3} \\ = & {(\frac{3}{13})}^{3} \\ = & 0.0123 \end{array}$

1.23%

$P (K, Q, J)$

$\begin{array}{rcl} P (K, Q, J) & = & {(\frac{4}{52})}^{3} = {(\frac{1}{13})}^{3} \\ = & 0.000455 \end{array}$

0.05%

For problems 12–16, use the following scenario.
A recent survey found that 75% of people left a positive review of a product. If eight customers are randomly selected, determine to the nearest hundredth of a percent, the chance of:

Exactly six people leaving a positive review

$\begin{array}{rcl} n & = & 8, r = 6, p = 0.75, q = 0.25 \\ P (6) & = & n C r (8, 6) \cdot {(0.75)}^{6} \cdot {(0.25)}^{8 - 6} \\ = & 0.3114 \end{array}$

31.14%

At least six people leaving a positive review

$n = 8, r = \{6, 7, 8\}, p = 0.75, q = 0.25 P (6) = nCr (8, 6) \cdot {(0.75)}^{6} \cdot {(0.25)}^{2} = 0.3115 P (7) = nCr (8, 7) \cdot {(0.75)}^{7} \cdot {(0.25)}^{1} = 0.2670 P (8) = nCr (8, 8) \cdot {(0.75)}^{8} \cdot {(0.25)}^{0} = 0.1001 P (6) + P (7) + P (8) = 0.6786$

67.86%

Note

In this lesson, we found P(6), P(7), and P(8) individually. However, if you put the entire problem in the calculator with the formula, you will get an answer of 67.85%.

Fewer than two people leaving a negative review

$n = 8, r = \{0, 1\}, p = 0.25, q = 0.75 P (0) = n C r (8, 0) \cdot {(0.25)}^{0} \cdot {(0.75)}^{8} = 0.1001 P (1) = nCr (8, 1) \cdot {(0.25)}^{1} \cdot {(0.75)}^{7} = 0.2670 P (0) + P (1) = 0.3671$

36.71%

All eight people leaving a positive review

$\begin{array}{rcl} n & = & 8, r = 8, p = 0.75, q = 0.25 \\ P (8) & = & n C r (8, 8) \cdot {(0.75)}^{8} \cdot {(0.25)}^{0} \\ = & 0.1001 \end{array}$

10.01%

If ten people were randomly selected, determine, to the nearest hundredth of a percent, the chance of exactly five people leaving a positive review.

$\begin{array}{rcl} n & = & 10, r = 5, p = 0.75, q = 0.25 \\ P (5) & = & n C r (10, 5) \cdot {(0.75)}^{5} \cdot {(0.25)}^{10 - 5} \\ = & 0.0584 \end{array}$

5.84%

Note

The number of trials changed.