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An experiment consists of tossing a coin and rolling a die. (a) Find the sample space. (b) Find the probability of getting heads and an even number. (c) Find the probability of getting heads and a number greater than 4. (d) Find the probability of getting tails and an odd number.

Short Answer

Expert verified
(a) Sample space: {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}. (b) \( \frac{1}{4} \). (c) \( \frac{1}{6} \). (d) \( \frac{1}{4} \).

Step by step solution

01

Identify possible outcomes of coin toss

The coin toss can result in two outcomes: Heads (H) or Tails (T).
02

Identify possible outcomes of die roll

The die roll can result in six outcomes: 1, 2, 3, 4, 5, or 6.
03

Construct the sample space

To construct the sample space, combine the coin outcomes with the die outcomes: (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6).
04

Calculate probability of heads and even number

Identify outcomes that have heads and an even number: (H,2), (H,4), (H,6). There are 3 such outcomes. There are 12 total outcomes, so the probability is \( \frac{3}{12} = \frac{1}{4} \).
05

Calculate probability of heads and number greater than 4

Identify outcomes that have heads and a number greater than 4: (H,5), (H,6). There are 2 such outcomes. The probability is \( \frac{2}{12} = \frac{1}{6} \).
06

Calculate probability of tails and odd number

Identify outcomes that have tails and an odd number: (T,1), (T,3), (T,5). There are 3 such outcomes. The probability is \( \frac{3}{12} = \frac{1}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Space
The concept of a sample space is central in probability theory. A sample space is essentially a list or set of all possible outcomes that could occur in a given experiment. Consider tossing a coin and rolling a die in one experiment. This experiment combines two smaller experiments: the coin toss and the die roll.

For the coin, there are two possibilities: Heads (H) or Tails (T). Meanwhile, the die can land on any of six numbers: 1 to 6. To list the outcomes of the experiment, we create combinations of the coin results and the die results. These combinations create the sample space for this experiment.

In this case, the sample space consists of the following pairs:
  • (H,1)
  • (H,2)
  • (H,3)
  • (H,4)
  • (H,5)
  • (H,6)
  • (T,1)
  • (T,2)
  • (T,3)
  • (T,4)
  • (T,5)
  • (T,6)
These 12 pairs encompass all possible results from this two-part experiment.
Coin Toss
A coin toss is one of the simplest probability experiments with an outcome that's always binary. This means that when we toss a coin, there are only two possible results: Heads (H) or Tails (T). This simplicity makes the coin toss a classic example in learning probability.

In our exercise, the coin toss is combined with the roll of a dice, creating a more varied sample space. Nonetheless, the basic idea remains unchanged: For each outcome of the die, you have a corresponding head or tail from the coin toss. This leads to pairs like (H,2) or (T,5), where the first part of the pair represents the coin toss result.
Die Roll
A die roll introduces more variability compared to a coin toss, with six possible outcomes as each face of the die is numbered from 1 to 6. This means there is a
  • 1/6 probability
of landing on any specific number when a fair die is rolled.

In our given problem, collaborating with a coin toss, it leads to a diverse sample space. For every result yielded from the coin toss, there are six potential results from the die. This interaction between the coin and die outcomes expands the range of overall possible outcomes in our experiment.
Outcomes
The idea of outcomes in probability refers to the result obtained from a random event or experiment. In the context of our problem, these are specific results like (H,3) or (T,6). Each combination is considered an outcome, and collectively, they define the sample space.

Identifying specific outcomes, such as heads with an even number, helps us answer probability questions related to our problem. For instance, an outcome like (H,2) or (T,1) relates directly to specific queries we might pose, such as the chance of rolling an odd number after getting tails. Thus, understanding the role of outcomes is vital in calculating probabilities accurately.

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Most popular questions from this chapter

Why is \(\left(\begin{array}{c}{n} \\ {r}\end{array}\right)\) the Same as \(C(n, r) ?\) This exercise explains why the binomial coefficients \(\left(\begin{array}{c}{n} \\ {r}\end{array}\right)\) that appear in the expansion of \((x+y)^{n}\) are the same as \(C(n, r),\) the number of ways of choosing \(r\) objects from \(n\) objects. First, note that expanding a binomial using only the Distributive Property gives $$\begin{aligned}(x+y)^{2} &=(x+y)(x+y) \\ &=(x+y) x+(x+y) y \\ &=x x+x y+y x+y y \end{aligned}$$ $$\begin{aligned}(x+y)^{3}=&(x+y)(x x+x y+y x+y y) \\\=& x x+x x y+x y x+x y y+y x x \\ &+y x y+y y x+y y y \end{aligned}$$ (a) Expand \((x+y)^{5}\) using only the Distributive Property. (b) Write all the terms that represent \(x^{2} y^{3}\) together. These are all the terms that contain two \(x^{\prime}\) s and three \(y^{\prime}\) s. (c) Note that two \(x^{\prime}\) s appear in all possible positions. Conclude that the number of terms that represent \(x^{2} y^{3}\) is \(C(5,2) .\) (d) In general, explain why \(\left(\begin{array}{c}{n} \\\ {r}\end{array}\right)\) in the Binomial Theorem is the same as \(C(n, r) .\)

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