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Chapter 16: Problem 17

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

Expert verified
(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

01

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.
02

Set up the sample space

Combine the possible outcomes of both dice. The sample space is: \ \((2,1), (2,2), (2,3), (4,1), (4,2), (4,3), (6,1), (6,2), (6,3)\).
03

Calculate the sums

Determine the sums for each pair of outcomes:\ \((2,1) \rightarrow 3,\ (2,2) \rightarrow 4,\ (2,3) \rightarrow 5,\ (4,1) \rightarrow 5,\ (4,2) \rightarrow 6,\ (4,3) \rightarrow 7,\ (6,1) \rightarrow 7,\ (6,2) \rightarrow 8,\ (6,3) \rightarrow 9\).
04

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.
05

Calculate the probabilities

Divide the frequency of each sum by the total number of outcomes (9): \ \(P(3) = \frac{1}{9}, \ P(4) = \frac{1}{9}, \ P(5) = \frac{2}{9}, \ P(6) = \frac{1}{9}, \ P(7) = \frac{2}{9}, \ P(8) = \frac{1}{9}, \ P(9) = \frac{1}{9}.\)
06

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 \ \(\frac{2}{9}\).
07

Calculate the probability of getting an even sum

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

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Key Concepts

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

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: \(\P(even) = \ \frac{1}{3}\).

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Most popular questions from this chapter

Some transistors of two different kinds (call them \(N\) and \(P)\) are stored in two boxes. You know that there are \(6 \mathrm{~N}^{\text {'s }}\) in one box and that \(2 \mathrm{~N}^{\prime} \mathrm{s}\) and \(3 \mathrm{P}^{\prime} \mathrm{s}\) got mixed in the other box, but you don't know which box is which, You select a box and a transistor from it at random and find that it is an \(N\); what is the probability that it came from the box with the \(6 \mathrm{~N}\) 's? From the other box? If another transistor is picked from the same box as the first, what is the probability that it is also an \(N\) ?

(a) A candy vending machine is out of order. The probability that you get a candy bar (with or without the return of your quarter) is \(\frac{1}{2}\), the probability that you get your quarter back (with or without candy) is \(\frac{1}{3}\), and the probability that you get both the candy and your money back is \(\frac{1}{12}\). What is the probability that you get nothing at all? Sugrestion: Sketch a geometric diagram similar to Figure \(3.1\), indicate regions representing the various possibilities and their probabilities; then set up a four-point samplc space and the associated probabilities of the points. (b) Suppose you. put another quarter into the candy vending machine of part (a). Set up the 16-point sample space corresponding to the possible results of your two attempts to bu? a candy bar, and find the probability that you get two candy bars (and no money back): that you get no candy and lose both quarters; that you just get your money back both times.

Five cards are dealt from a shuffled deck. What is the probability that they are all of the same suit? That they are all diamonds? That they are all face cards? That the five cards are a sequence in the same suit (for example, \(3,4,5,6,7\) of hearts)?

(a) Find the probability that in two tosses of a coin, one is heads and one tails. That in six tosses of a die, all six of the faces show up. That in 12 tosses of a 12-sided die, all 12 faces show up. That in \(n\) tosses of an \(n\)-sided die, all \(n\) faces show up. (b) The last problem in (a) is equivalent to finding the probability that, when \(n\) balls are distributed at random into \(n\) boxes, each box contains exacty one ball. Show that for large \(n\), this is approximateiy \(e^{-n} \sqrt{2 \pi n}\)

Suppose you have 3 nickels and 4 dimes in your right pocket and 2 nickels and a quarter in your left pocket. You pick a pocket at random and from it select a coin at random. If it is a. nickel, what is the probability that it came from your right pocket?
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