Problem 1
Question
Adice is thrown twice and the sum of the numbers appearing is observed to be 6 . What is the conditional probability that the number 4 has appeared at least once?
Step-by-Step Solution
Verified Answer
The conditional probability is \( \frac{2}{5} \).
1Step 1: Identify Possible Outcomes
A single die can result in one of 6 outcomes: 1, 2, 3, 4, 5, or 6. When two dice are thrown, there are a total of 6 x 6 = 36 possible outcomes.
2Step 2: Calculate Sum of 6 Outcomes
List the outcomes where the sum of the numbers is 6. These outcomes are: (1,5), (2,4), (3,3), (4,2), and (5,1). Thus, there are 5 favorable outcomes when the sum is 6.
3Step 3: Identify Favorable Outcomes
Since 4 needs to appear at least once, identify the outcomes from Step 2 that include at least one 4. The favorable pairs here are: (2,4) and (4,2).
4Step 4: Calculate Conditional Probability
The probability of event A (4 appearing at least once when the sum is 6) is the ratio of favorable outcomes with a 4 to total outcomes with a sum of 6. Therefore, the conditional probability is \( \frac{2}{5} \).
Key Concepts
Dice ProbabilitiesProbability TheoryOutcomes in Probability
Dice Probabilities
Dice probabilities are a key concept when dealing with situations involving dice, such as the exercise about a die thrown twice. When we consider just one die, there are 6 possible outcomes: 1 through 6.
Throwing two dice expands this to a total of 36 possible outcomes, since each die operates independently, and thus, any number appearing on the first die can be coupled with any number on the second. As a result, to determine the probability of any specific outcome or combination of outcomes, we use the basic rule of probability: the number of favorable outcomes divided by the total number of possible outcomes.
Throwing two dice expands this to a total of 36 possible outcomes, since each die operates independently, and thus, any number appearing on the first die can be coupled with any number on the second. As a result, to determine the probability of any specific outcome or combination of outcomes, we use the basic rule of probability: the number of favorable outcomes divided by the total number of possible outcomes.
- For a single die: 6 outcomes (1-6).
- For two dice: 36 outcomes (each die can be any of 1-6, so 6 possibilities for the first die times 6 for the second).
Probability Theory
Probability theory is the mathematical study of randomness and uncertainty. It provides the tools to model and reason about situations that involve uncertainty, such as throwing a die. Probability tells us how likely it is that a particular event will happen. Events can be seen on a spectrum from certainty (happening 100% of the time) to impossibility (never happening).
In the context of our dice example:
In the context of our dice example:
- Each outcome (combination of numbers rolled) has a probability of occurring determined by dividing the number of ways to achieve that outcome by the total possible outcomes.
- Conditional probability is a subpart of probability theory that considers the likelihood of an event, given that another event has already occurred. This is exactly what we solve for when determining the probability of rolling a 4 when the sum is known to be 6.
Outcomes in Probability
Outcomes in probability refer to the potential results we get from an experiment. In the dice example, an outcome would be the result of a roll of two dice.
To solve the given problem of finding conditional probability, we first identify all possible outcomes that meet the condition of interest, such as the sum being 6. Once identified, the task is to further filter those to include only the outcomes where a 4 appears at least once.
To solve the given problem of finding conditional probability, we first identify all possible outcomes that meet the condition of interest, such as the sum being 6. Once identified, the task is to further filter those to include only the outcomes where a 4 appears at least once.
- The possible sums of 6 are: (1,5), (2,4), (3,3), (4,2), and (5,1).
- Favorable outcomes where a 4 appears are: (2,4) and (4,2).
Other exercises in this chapter
Problem 1
Seven chits are numbered 1 to 7 . Three are drawn one by one with replacement. The probability that the least number on any selected chit is 5 is (a) \(1-\left(
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If \(4 P(A)=6 P(B)=10(A \cap B)=1\), then \(P(B / \mathrm{A})=\) (a) \(2 / 5\) (b) \(3 / 5\) (c) \(7 / 10\) (d) \(19 / 60\)
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A box contains 2 black, 4 white and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is d
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