Problem 62
Question
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007] (a) \(8 / 729\) (b) \(8 / 243\) (c) \(1 / 729\) (d) \(8 / 9\).
Step-by-Step Solution
Verified Answer
(b) \(8 / 243\)
1Step 1: Calculate Probability of Getting a Score of Exactly 9 Once
The score of 9 can be obtained from a pair of dice with the following combinations: (3,6), (4,5), (5,4), and (6,3). There are a total of 4 ways to obtain this score. Since there are 36 possible outcomes when throwing two dice, the probability of getting a score of 9 in one throw is given by: \[P( ext{Score of 9}) = \frac{4}{36} = \frac{1}{9}.\]
2Step 2: Probability of Not Getting a Score of 9
If the probability of obtaining exactly 9 is \( \frac{1}{9} \), the probability of not getting a score of 9 is:\[P( ext{Not 9}) = 1 - \frac{1}{9} = \frac{8}{9}.\]
3Step 3: Use the Binomial Probability Formula
To find the probability of getting exactly two scores of 9 in three throws, we apply the binomial probability formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k},\]where \(n = 3\) (number of trials), \(k = 2\) (number of successes), and \(p = \frac{1}{9}\) (probability of success in one trial).
4Step 4: Substitute Values into the Formula
Substituting the values into the binomial formula, we have:\[P(X = 2) = \binom{3}{2} \left(\frac{1}{9}\right)^2 \left(\frac{8}{9}\right)^1\]\[= 3 \times \frac{1}{81} \times \frac{8}{9}\]\[= 3 \times \frac{8}{729}\]\[= \frac{24}{729}.\]
5Step 5: Simplify the Fraction
Simplify the fraction \(\frac{24}{729}\):\[\frac{24}{729} = \frac{8}{243}.\]
Key Concepts
Probability TheoryCombinatoricsDice Probability
Probability Theory
Probability refers to the measure of the likelihood that an event will occur. In probability theory, every possible outcome of a random event is assigned a probability, ranging between 0 and 1.
For instance, an event with probability 0 will not occur, whereas an event with probability 1 is certain to occur. When dealing with multiple events, the probabilities of all possible outcomes add up to 1.
To understand probability, consider scenarios like rolling dice, where each side (or face) has an equal chance of occurring. By knowing each event's probability, we can make informed predictions and decisions.
In situations involving repeated independent trials, as in the exercise where dice are rolled multiple times, probability becomes a valuable tool for understanding potential outcomes. It's crucial to differentiate between independent and dependent events. In our example, each dice roll is independent, meaning the result of one roll doesn't affect others.
For instance, an event with probability 0 will not occur, whereas an event with probability 1 is certain to occur. When dealing with multiple events, the probabilities of all possible outcomes add up to 1.
To understand probability, consider scenarios like rolling dice, where each side (or face) has an equal chance of occurring. By knowing each event's probability, we can make informed predictions and decisions.
In situations involving repeated independent trials, as in the exercise where dice are rolled multiple times, probability becomes a valuable tool for understanding potential outcomes. It's crucial to differentiate between independent and dependent events. In our example, each dice roll is independent, meaning the result of one roll doesn't affect others.
Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. It underpins many probability problems such as the one in the exercise, helping to simplify complex calculations.
In our exercise, finding how many ways we can roll a specific sum with a pair of dice, like a 9, relies on combinatorics. We have possible outcomes like (3,6), (4,5), (5,4), and (6,3). Counting these combinations is essential to calculate probabilities accurately.
For more complex problems, the concept of combinations is used, denoted by binomial coefficients \(\binom{n}{k}\), which represent the number of ways to choose \(k\) successes out of \(n\) trials. This is evident when using the binomial probability formula to determine the probability of getting exactly two scores of 9 in three throws of a dice. Combinatorics provides a systematic way to organize and count outcomes.
In our exercise, finding how many ways we can roll a specific sum with a pair of dice, like a 9, relies on combinatorics. We have possible outcomes like (3,6), (4,5), (5,4), and (6,3). Counting these combinations is essential to calculate probabilities accurately.
For more complex problems, the concept of combinations is used, denoted by binomial coefficients \(\binom{n}{k}\), which represent the number of ways to choose \(k\) successes out of \(n\) trials. This is evident when using the binomial probability formula to determine the probability of getting exactly two scores of 9 in three throws of a dice. Combinatorics provides a systematic way to organize and count outcomes.
Dice Probability
Dice probability is a specific application of probability theory that deals with the outcomes of rolling dice. A fair die has six faces, each equally likely to appear when rolled.
In a standard pair of dice, each face of the first die is independent of the second. This creates a total of 36 possible outcomes when both are rolled. Among these combinations, certain totals are more common than others.
For instance, obtaining a total score of 9 can happen in four distinct ways: through the pairs (3,6), (4,5), (5,4), and (6,3). This insight allows us to calculate the probability of rolling a nine on a single throw as \(\frac{4}{36} = \frac{1}{9}\).
When faced with more complicated problems, such as determining the probability of achieving a score of 9 twice in three throws, we use advanced techniques like binomial probability. By understanding these principles, we gain a deeper appreciation of how outcomes are determined in games of chance.
In a standard pair of dice, each face of the first die is independent of the second. This creates a total of 36 possible outcomes when both are rolled. Among these combinations, certain totals are more common than others.
For instance, obtaining a total score of 9 can happen in four distinct ways: through the pairs (3,6), (4,5), (5,4), and (6,3). This insight allows us to calculate the probability of rolling a nine on a single throw as \(\frac{4}{36} = \frac{1}{9}\).
When faced with more complicated problems, such as determining the probability of achieving a score of 9 twice in three throws, we use advanced techniques like binomial probability. By understanding these principles, we gain a deeper appreciation of how outcomes are determined in games of chance.
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