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A fair die with sides numbered \(1,2,3,4,5,\) and 6 is to be rolled 4 times. What is the probability that on at least one roll the number showing will be less than \(3 ?\) a. \(\frac{65}{81}\) b. \(\frac{67}{81}\) c. \(\frac{8}{9}\) d. \(\frac{26}{27}\) e. \(\frac{80}{81}\)

Short Answer

Expert verified
The probability is \( \frac{65}{81} \).

Step by step solution

01

- Define the Events

Identify the event of interest. Event A is when the number showing is less than 3 on at least one roll.
02

- Calculate the Probability of the Complement Event

Find the probability of all rolls showing a number of 3 or more. There are 4 dice rolls, and each die has 4 sides (3, 4, 5, 6) that are 3 or more. Hence, the probability of one die showing a number 3 or more is \( P(B) = \frac{4}{6} = \frac{2}{3} \).
03

- Compute the Probability of All Rolls Showing a Number of 3 or More

The complement of event A happening on all four rolls is \( P(B_{all}) = (\frac{2}{3})^4 = \frac{16}{81} \).
04

- Find the Probability of Event A

Subtract the probability found in Step 3 from 1 to get the probability of at least one roll showing less than 3: \( P(A) = 1 - P(B_{all}) = 1 - \frac{16}{81} = \frac{65}{81} \).
05

- Verify the Answer Choice

Compare the result with given answer choices to confirm the correct answer. The correct answer matches option a.

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Key Concepts

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

Dice Probability
Understanding probability with dice is a fundamental concept in probability theory. A standard die has six faces, each showing a different number from 1 to 6. Each number has an equal chance of landing face up when the die is rolled.

When we discuss the probability of events related to dice rolls, we use the principle that each outcome is equally likely. For instance, the probability of rolling a number less than 3 on a single die is calculated by counting the favorable outcomes (1 and 2) and dividing by the total number of possible outcomes (6), which gives us:
\[ P(\text{less than 3}) = \frac{2}{6} = \frac{1}{3} \]
To solve problems involving multiple rolls of the die, we extend these basic principles. This means calculating probabilities for each roll and combining them appropriately.
Complement Rule
The complement rule is a useful tool in probability, especially when dealing with complicated events. The complement of an event A, denoted as A', consists of all outcomes that are not in A. The sum of the probabilities of an event and its complement is always 1. Mathematically, it's expressed as:
\[ P(A') = 1 - P(A) \]
In our exercise, instead of directly calculating the probability of rolling a number less than 3 on at least one of 4 rolls, we use the complement rule. We first find the probability of the opposite event, which is all four rolls showing numbers 3 or greater. Once we have this probability, we subtract it from 1 to find the required probability.

By recognizing that calculating the complement is simpler, our problem-solving becomes more efficient without changing the final outcome. This approach is especially helpful when dealing with probabilities of multiple events.
Multiple Events Probability
When dealing with multiple independent events, such as rolling a die multiple times, we often need to calculate the combined probability. If events are independent, the probability of multiple events all occurring is the product of their individual probabilities.

In our die example, each roll is independent, meaning the result of one roll doesn’t impact the others. To find the probability that all four rolls show a number 3 or more, we calculate the probability for a single roll and then raise it to the power of the number of rolls:
\[ P(\text{all rolls } \geq 3) = \left( \frac{2}{3} \right)^4 = \frac{16}{81} \]

It's always critical to verify whether events are indeed independent before applying these rules. Misidentifying dependencies can lead to incorrect probabilities, causing errors in problem-solving and conclusions.

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Most popular questions from this chapter

If \(\left(x^{2}+8\right) y z<0, w z>0,\) and \(x y z<0,\) the which of the following must be true? I. \(x<0\) II. \(w y<0\) III. \(y z<0\) a. II only b. III only c. I and III only d. II and III only e. I, II, and III

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