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What is the probability of rolling two fair six-sided dice a. and getting a sum greater than or equal to \(7 ?\) b. getting an even sum or a sum greater than \(7 ?\)

Short Answer

Expert verified
a. \( \frac{7}{12} \); b. \( \frac{2}{3} \).

Step by step solution

01

Calculate Total Possible Outcomes

When rolling two six-sided dice, each die has 6 faces, so the total number of possible outcomes is calculated by multiplying the outcomes of each die: \( 6 \times 6 = 36 \). Thus, there are 36 possible outcomes in total for rolling two dice.
02

Determine Outcomes for Sum Greater or Equal to 7

Calculate the number of outcomes where the sum of the dice is greater than or equal to 7. The possible sums and their outcomes are: 7: \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\) (6 outcomes), 8: \((2,6), (3,5), (4,4), (5,3), (6,2)\) (5 outcomes), 9: \((3,6), (4,5), (5,4), (6,3)\) (4 outcomes), 10: \((4,6), (5,5), (6,4)\) (3 outcomes), 11: \((5,6), (6,5)\) (2 outcomes), 12: \((6,6)\) (1 outcome). Adding them gives 21 outcomes.
03

Calculate Probability for Part (a)

The probability is the number of favorable outcomes divided by the total number of possible outcomes: \( \frac{21}{36} = \frac{7}{12} \).
04

Determine Outcomes for Even Sum

Find the outcomes for which the sum of the dice is even: possible even sums are 2, 4, 6, 8, 10, and 12. These sums give us outcomes for the number of even pairs: 2: \((1,1)\) (1 outcome), 4: \((1,3), (2,2), (3,1)\) (3 outcomes), 6: \((1,5), (2,4), (3,3), (4,2), (5,1)\) (5 outcomes), 8: \((2,6), (3,5), (4,4), (5,3), (6,2)\) (5 outcomes), 10: \((4,6), (5,5), (6,4)\) (3 outcomes), 12: \((6,6)\) (1 outcome). Adding all gives 18 outcomes.
05

Determine Outcomes for Sum Greater Than 7

Outcomes with sum greater than 7 are those for sums 8, 9, 10, 11, and 12, which we found previously total 15 (from part a without 7).
06

Calculate Probability for Part (b) by Inclusion-Exclusion Principle

Use the inclusion-exclusion principle to find the probability of either an even sum or a sum greater than 7. Calculate for even outcomes: 18 outcomes, greater than 7: 15 outcomes. Sum overlaps: since every number greater than 7 in this list 8, 10, 12 is even, overlap is 9 (from earlier sum of 8, 10, and 12). Calculate: \(18 + 15 - 9 = 24\). The probability is \( \frac{24}{36} = \frac{2}{3} \).

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Key Concepts

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

Probability with Dice
Dice are one of the most common tools used to explain probability. Each die has six faces numbered from 1 to 6. When rolling two six-sided dice, you calculate the possible outcomes by multiplying the possibilities of each die. Thus, for two dice, there are 36 possible combinations \(6 \times 6 = 36\).
  • When calculating probabilities, we ask how many favorable outcomes meet the conditions of the question.
  • The probability of a specific event occurring is given by dividing the number of favorable outcomes by the total possible outcomes.
  • For example, to find the probability of getting a certain sum, say 7, we look at all pairs that add up to 7, compare it to the total combinations, and calculate.
Understanding these basics helps break down more complex probability problems involving dice.
Even Sum
The term 'even sum' means that the total of the two numbers rolled on the dice is an even number. On a six-sided die, the possible sums that are even are 2, 4, 6, 8, 10, and 12. Each of these can be achieved by different combinations of rolls:
  • For a sum of 2, the roll must be (1, 1).
  • For a sum of 4, we can roll (1, 3), (2, 2), or (3, 1).
  • A sum of 6 can be achieved with (1, 5), (2, 4), (3, 3), (4, 2), or (5, 1).
  • And so forth for the other even sums.
Summing these up gives 18 out of the 36 possible outcomes, reflecting how often we might see an even sum. This division of outcomes into even sums helps us understand and compute probabilities more accurately.
Inclusion-Exclusion Principle
In probability, the inclusion-exclusion principle allows us to calculate the probability of the union of two sets. When rolling dice, it helps in determining how often we get either one condition or another.
  • For two events A and B, the principle states: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
  • The overlap \( P(A \cap B) \) represents the probability of both events occurring simultaneously and needs to be subtracted to avoid double-counting.
In this case, to find the probability of rolling an even sum or a sum greater than 7, we use this principle. We compute the individual probabilities of each event, calculate their overlap, and then apply the inclusion-exclusion principle to get: 18 outcomes for even sums plus 15 outcomes for sums greater than 7, minus 9 overlapping outcomes (for sums like 8, 10, and 12), resulting in 24 favorable outcomes. The probability then is \( \frac{24}{36} = \frac{2}{3} \). This approach helps in accurately capturing the true count of favorable outcomes.

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Most popular questions from this chapter

The contingency table below shows the survival data for the passengers of the Titanic. $$ \begin{array}{|c|c|c|c|c|c|} \hline & \text { First } & \text { Second } & \text { Third } & \text { Crew } & \text { Total } \\ \hline \text { Survive } & 203 & 118 & 178 & 212 & 711 \\ \hline \text { Not Survive } & 122 & 167 & 528 & 673 & 1490 \\ \hline \text { Total } & 325 & 285 & 706 & 885 & 2201 \\ \hline \end{array} $$ a. What is the probability that a passenger did not survive? b. What is the probability that a passenger was crew? c. What is the probability that a passenger was first class and did not survive? d. What is the probability that a passenger did not survive or was crew? e. What is the probability that a passenger survived given they were first class? f. What is the probability that a passenger survived given they were second class? g. What is the probability that a passenger survived given they were third class? h. Does it appear that survival depended on the passenger's class? Or are they independent? Use probability to support your claim.

A fair six-sided die is rolled twice. What is the probability of getting: a. a 6 on both rolls? b. a 5 on the first roll and an even number on the second roll?

A ball is drawn randomly from a jar containing 6 red marbles, 2 white marbles, and 5 yellow marbles. Find the probability of: a. Drawing a white marble. b. Drawing a red marble. c. Drawing a green marble. d. Drawing two yellow marbles if you draw with replacement. e. Drawing first a red marble then a white marble if marbles are drawn without replacement.

A box contains four black pieces of cloth, two striped pieces, and six dotted pieces. A piece is selected randomly and then placed back in the box. A second piece is selected randomly. What is the probability that: a. both pieces are dotted? b. the first piece is black, and the second piece is dotted? c. one piece is black, and one piece is striped?

A professor gave a test to students in a morning class and the same test to the afternoon class. The grades are summarized below. $$ \begin{array}{|c|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \text { Total } \\ \hline \text { Morning Class } & 8 & 18 & 13 & 39 \\ \hline \text { Afternoon Class } & 10 & 4 & 12 & 26 \\ \hline \text { Total } & 18 & 22 & 25 & 65 \\ \hline \end{array} $$ If one student was chosen at random: a. What is the probability they were in the morning class? b. What is the probability they earned a C? c. What is the probability that they earned an \(A\) and they were in the afternoon class? d. What is the probability that they earned an A given they were in the morning class? e. What is the probability that they were in the morning class or they earned a \(\mathrm{B} ?\)

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