Question

Two dice are thrown. Given the information that the number on the first die is even, and the number on the second is \(<4\), set up an appropriate sample space and answer the following questions. (a) What are the possible sums and their probabilities? (b) What is the most probable sum? (c) What is the probability that the sum is even?

Short answer

(a) Possible sums and their probabilities are: 3: 1/9, 4: 1/9, 5: 2/9, 6: 1/9, 7: 2/9, 8: 1/9, 9: 1/9. (b) Most probable sums are 5 and 7. (c) Probability of even sum is 1/3.

Step-by-step solution

Identify the constraints

The first die shows an even number, and the second die shows a number less than 4. The possible outcomes for the first die are 2, 4, and 6. The possible outcomes for the second die are 1, 2, and 3.

Set up the sample space

Combine the possible outcomes of both dice. The sample space is: \ \((2,1), (2,2), (2,3), (4,1), (4,2), (4,3), (6,1), (6,2), (6,3)\).

Calculate the sums

Determine the sums for each pair of outcomes:\ \((2,1) \rightarrow 3,\ (2,2) \rightarrow 4,\ (2,3) \rightarrow 5,\ (4,1) \rightarrow 5,\ (4,2) \rightarrow 6,\ (4,3) \rightarrow 7,\ (6,1) \rightarrow 7,\ (6,2) \rightarrow 8,\ (6,3) \rightarrow 9\).

Count the frequencies of each sum

Tally the frequency of each sum: \ \(Sum = 3\), Frequency = 1; \(Sum = 4\), Frequency = 1; \(Sum = 5\), Frequency = 2; \(Sum = 6\), Frequency = 1; \(Sum = 7\), Frequency = 2; \(Sum = 8\), Frequency = 1; \(Sum = 9\), Frequency = 1.

Calculate the probabilities

Divide the frequency of each sum by the total number of outcomes (9): \ \(P(3) = \frac{1}{9}, \ P(4) = \frac{1}{9}, \ P(5) = \frac{2}{9}, \ P(6) = \frac{1}{9}, \ P(7) = \frac{2}{9}, \ P(8) = \frac{1}{9}, \ P(9) = \frac{1}{9}.\)

Identify the most probable sum

The most probable sums are the ones with the highest probability. Here, sums of 5 and 7 have the highest probability, each occurring \ \(\frac{2}{9}\).

Calculate the probability of getting an even sum

Sum the probabilities of each even sum: \ \ \ \(P(even) = P(4) + P(6) + P(8) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}.\)

Key concepts

Sample Space

To understand the probabilities involved when throwing two dice, we first need to define the sample space. The sample space represents all possible outcomes of an experiment.
In this exercise, the first die must show an even number, and the second die must show a number less than 4.
The possible outcomes for the first die are 2, 4, and 6. For the second die, the potential outcomes are 1, 2, and 3.
By combining these, we get the sample space: \((2,1), (2,2), (2,3), (4,1), (4,2), (4,3), (6,1), (6,2), (6,3)\). These pairs represent all possible results of rolling two dice under the given conditions. Each pair of numbers (first die, second die) is an element of this sample space.

Probability Calculation

Once we have the sample space, we can calculate the probabilities for different outcomes. The total number of outcomes in our sample space is 9.
Probability is calculated by dividing the number of favorable outcomes by the total number of outcomes.
For example, to find the probability of a specific outcome like a sum of 5, we first determine how many pairs in our sample space produce this sum.
Pairs \((2,3)\) and \((4,1)\) give us a sum of 5. Since there are 2 pairs that total 5, the probability is \P(5) = \frac{2}{9}\.

Sum of Dice

When rolling two dice, each pair of outcomes can be added to get a sum. For instance, the pairs \((2,1)\) and \((6,3)\) have sums of 3 and 9, respectively.
The sums for all pairs in our sample space are:

• \Sum = 3\, Frequency = 1d
• \Sum = 4\, Frequency = 1
• \Sum = 5\, Frequency = 2d
• \Sum = 6\, Frequency = 1
• \Sum = 7\, Frequency = 2
• \Sum = 8\, Frequency = 1
• \Sum = 9\, Frequency = 1

We can see that sums of 5 and 7 appear most frequently.

Frequency and Probability Distribution

Frequency tells us how often each sum occurs in our sample space. From frequency, we derive the probability distribution, which shows the chance of each sum.
The frequency distribution in this problem is:
\(\ P(3) = \frac{1}{9}, \ P(4) = \frac{1}{9}, \ P(5) = \frac{2}{9}, \ P(6) = \frac{1}{9}, \ P(7) = \frac{2}{9}, \ P(8) = \frac{1}{9}, \ P(9) = \frac{1}{9}. \)
This distribution helps us easily identify the most probable sums, which in this case are 5 and 7, both having a probability of \(\ \frac{2}{9}\).
Lastly, to find the probability of getting an even sum, we sum the probabilities of sums 4, 6, and 8: \(\P(even) = \ \frac{1}{3}\).