Given the sample space S = {(H, 1), (T, 1), (H, 2), (T, 2), (H, 3), (T, 3), (H, 4), (T, 4), (H, 5), (T, 5), (H, 6), (T, 6)}, determine the probabilities of specified events using c... Given the sample space S = {(H, 1), (T, 1), (H, 2), (T, 2), (H, 3), (T, 3), (H, 4), (T, 4), (H, 5), (T, 5), (H, 6), (T, 6)}, determine the probabilities of specified events using commas to separate multiple inputs, and enter EMPTY if the event is an impossible event.
Understand the Problem
The question provides a sample space S consisting of outcomes from tossing a fair coin (Heads or Tails) and rolling a fair die (numbers 1-6). Each outcome is an ordered pair (coin result, die result). The question requires you to use this sample space to determine probabilities of events. The prompt indicates to enter 'EMPTY' if an impossible event occurs, and use commas to separate answers for multiple inputs.
Answer
$\frac{1}{12}, \frac{1}{4}, \frac{1}{6}, \text{EMPTY}$
Answer for screen readers
$\frac{1}{12}, \frac{1}{4}, \frac{1}{6}, \text{EMPTY}$
Steps to Solve
- List the Sample Space
The sample space $S$ consists of all possible outcomes of tossing a coin and rolling a die. We can represent these outcomes as ordered pairs $(C, D)$, where $C$ is either H (Heads) or T (Tails), and $D$ is a number from 1 to 6. Listing all possible outcomes:
$S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}$
Since the coin and die are fair, each outcome in the sample space is equally likely. The total number of outcomes is 12.
- Determine $P(H, 2)$
The event $(H, 2)$ is a single outcome in the sample space. Since there are 12 equally likely outcomes, the probability of $(H, 2)$ is $\frac{1}{12}$.
- Determine $P(\text{Tails and an even number})$
The event "Tails and an even number" corresponds to the outcomes $(T, 2)$, $(T, 4)$, and $(T, 6)$. There are 3 favorable outcomes. Since there are 12 total outcomes, the probability is $\frac{3}{12} = \frac{1}{4}$.
- Determine $P(\text{Heads and a number greater than 4})$
The event "Heads and a number greater than 4" corresponds to the outcomes $(H, 5)$ and $(H, 6)$. There are 2 favorable outcomes. Since there are 12 total outcomes, the probability is $\frac{2}{12} = \frac{1}{6}$.
- Determine $P(\text{Tails and a 7})$
The event "Tails and a 7" is impossible since the die only has numbers from 1 to 6. Therefore, the probability is 0. Since the prompt indicates to enter 'EMPTY' for an impossible event, we enter 'EMPTY'.
- List the probabilities
Listing the probabilities with commas, we have $\frac{1}{12}, \frac{1}{4}, \frac{1}{6}, \text{EMPTY}$.
$\frac{1}{12}, \frac{1}{4}, \frac{1}{6}, \text{EMPTY}$
More Information
The individual probabilities could be expressed as decimals, but the fractional representation is more exact.
The question tests basic probability concepts within a well-defined sample space.
Tips
- Forgetting to consider all the possibilities for outcomes, especially when defining the sample space or a specific event.
- Not simplifying fractions to their simplest form. Although $\frac{3}{12}$ is technically correct, $\frac{1}{4}$ is better.
- Incorrectly identifying impossible events and not entering 'EMPTY' when required.
- Calculating probability as number of total outcomes divided by number of favorable outcomes instead of the other way around.
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