Practice 1 Solutions

For problems 1–3, use the scenario.

For a standard deck of 52 cards, a student draws cards randomly, without replacement. Determine the probability. Write your answer as a fraction.

$P (K, Q, 2)$

$\begin{array}{rcl} P (K, Q, 2) & = & \frac{4}{52} \cdot \frac{4}{51} \cdot \frac{4}{50} \\ = & \frac{8}{16575} \end{array}$

$\begin{array}{rcl} \frac{8}{16575} \end{array}$

$P (spade, spade, heart)$

$\begin{array}{rcl} P (spade, spade, heart) & = & \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{13}{50} \\ = & \frac{13}{850} \end{array}$

$\begin{array}{rcl} \frac{13}{850} \end{array}$

$P (7, 7, 7)$

$\begin{array}{rcl} P (7, 7, 7) & = & \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \\ = & \frac{1}{5525} \end{array}$

$\begin{array}{rcl} \frac{1}{5525} \end{array}$

For problems 4–6, use the scenario.

A room consists of three 9th graders, five 10th graders, eight 11th graders, and four 12th graders. Determine the percent chance of students being randomly selected into groups of:

Two ninth graders, one tenth grader, and one eleventh grader

$\begin{array}{rcl} P (9 th, 9 th, 10 th, 11 th) & = & \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{5}{18} \cdot \frac{8}{17} \\ = & \frac{2}{969} = 0.0021 \end{array}$

$\begin{array}{rcl} 0.21 % \end{array}$

One twelfth grader and one eleventh grader

$\begin{array}{rcl} P (12 th, 11 th) & = & \frac{4}{20} \cdot \frac{8}{19} \\ = & \frac{8}{95} = 0.0842 \end{array}$

$8.42 %$

One ninth grader, one tenth grader, one twelfth grader

$\begin{array}{rcl} P (9 th, 10 th, 12 th) & = & \frac{3}{20} \cdot \frac{5}{19} \cdot \frac{4}{18} \\ = & \frac{1}{114} = 0.0088 \end{array}$

$\begin{array}{rcl} 0.88 % \end{array}$

For problems 7–11, use the scenario.

In the Midwest, there is a 0.7 probability of a sunny day during the 100 days of summer. If it is sunny, there is a 0.2 probability of a thunderstorm in the afternoon. If it is raining, the probability of a thunderstorm in the afternoon is 0.6.

S: Sunny day 
R: Rain 
T: Thunderstorm

$P (S) = 0.7 P (R) = 1 - P (S) = 0.3 P (T | S) = 0.2 P (T | R) = 0.6$

Note

Q: How many days of summer are there according to the given information?

A: 100 days

Q: Since percent means out of 100, how will this be helpful in determining the answers for problems 7–11?

A: The decimal values can be multiplied by 100 (or the decimal point can be moved two spaces to the right).

Determine the number of summer days that are sunny, AND there is a thunderstorm.

$\begin{array}{rcl} P (S \cap T) & = & P (S) \cdot P (T ∣ S) \\ = & 0.7 \cdot 0.2 \\ = & 0.14 \end{array}$

14 days

Note

This is $P (S \cap T) .$

Determine the total number of summer days a thunderstorm can be expected.

$P (S \cap T) = P (S) \cdot P (T | S) = 0.7 \cdot 0.2 = 0.14 \begin{array}{rcl} P (R \cap T) & = & P (S) \cdot P (T ∣ R) = 0.3 \cdot 0.6 = 0.18 \\ P (T) & = & P (S \cap T) + P (R \cap T) \\ P (T) & = & 0.14 + 0.18 = 0.32 \end{array}$

32 days

Note

A thunderstorm can happen if it is sunny AND there is a thunderstorm, OR if it is raining AND there is a thunderstorm.

Determine the number of days in the summer that are sunny, AND there is NO thunderstorm.

$\begin{array}{rcl} P (T ‘) & = & 1 - 0.2 = 0.8 \\ P (S \cap T ‘) & = & P (S) \cdot P (T ‘ | S) \\ = & 0.7 \cdot 0.8 \\ = & 0.56 \end{array}$

56 days

Note

To find the probability of both a sunny day and no thunderstorm, we multiply the probability of a sunny day by the conditional probability of no thunderstorm given it is sunny.

On how many summer days will there likely be rain and NO thunderstorm?

$\begin{array}{rcl} P (R \cap T ‘) = 1 - 0.6 & = & 0.4 \\ P (R \cap T ‘) & = & P (R) \cdot P (T ‘ | R) \\ = & 0.3 \cdot 0.4 \\ = & 0.12 \end{array}$

12 days

Note

Since the probability of a thunderstorm on a not sunny day is 0.6, the probability of no thunderstorm on a rainy day is $1 - 0.6 = 0.4 .$

How many summer days will likely have no thunderstorms?

$\begin{array}{rcl} P (T ‘) & = & P (S \cap T ‘) + P (R \cap T ‘) \\ = & 0.56 + 0.12 \\ = & 0.68 \end{array}$

68 days

Note

These are mutually exclusive events, meaning they cannot happen at the same time, so the probabilities can be added together.

For problems 12–14, use the scenario.

At a local gym, 60% of the members use the cardio machines while 45% of the members lift weights. The probability that a member uses the cardio machines, given they lift weights, is 80%.

Explain why the events are not independent.

When $P (A | B) = P (A)$ , the events A and B are independent.

$(0.60) (0.45) \neq 0.80$

Because $P (C | W) \neq P (C)$ , the events are not independent.

What is the percent chance that a randomly selected member uses the cardio machines and lifts weights?

$\begin{array}{rcl} P (C \cap W) & = & P (C | W) \cdot P (W) \\ = & 0.80 \cdot 0.45 \\ = & 0.36 \end{array}$

36%

$P (W | C)$

$\begin{array}{rcl} P (W | C) & = & \frac{P (C \cap W)}{P (C)} \\ = & \frac{0.36}{0.60} \\ = & 0.60 \end{array}$

60%