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If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is \(i\) ? Compute for all values of \(i\) between 2 and 12 .

Short Answer

Expert verified
The conditional probabilities for the first die landing on 6 given that the sum of the dice is \(i\) are: \(P(first\_die\_6 | sum\_2) = 0\) \(P(first\_die\_6 | sum\_3) = 0\) \(P(first\_die\_6 | sum\_4) = 0\) \(P(first\_die\_6 | sum\_5) = 0\) \(P(first\_die\_6 | sum\_6) = \frac{1}{5}\) \(P(first\_die\_6 | sum\_7) = \frac{1}{6}\) \(P(first\_die\_6 | sum\_8) = \frac{1}{5}\) \(P(first\_die\_6 | sum\_9) = \frac{1}{4}\) \(P(first\_die\_6 | sum\_10) = \frac{1}{3}\) \(P(first\_die\_6 | sum\_11) = \frac{1}{2}\) \(P(first\_die\_6 | sum\_12) = 1\)

Step by step solution

01

Recall the conditional probability formula

The conditional probability formula is given by: \(P(A | B) = \frac{P(A \cap B)}{P(B)}\) where \(P(A | B)\) is the probability of event A happening given that event B happens, \(P(A \cap B)\) is the probability of both A and B happening, and P(B) is the probability of event B happening.
02

Find the sample space for the sum of two dice

When two dice are rolled, there are a total of 36 possible outcomes (6 sides on the first die and 6 sides on the second die). We can determine the total number of outcomes resulting in a sum of \(i\) for each value of \(i\) between 2 and 12.
03

Find the probability of the first die landing on 6 and the sum being i

For each value of \(i\) (2 to 12), we will calculate the probability of the sum being \(i\) and the first die landing on 6.
04

Calculate the conditional probability with the given information

For each value of \(i\) (2 to 12), using the conditional probability formula, we will calculate the probability that the first die lands on 6 given that the sum of the dice is \(i\). Here are the calculations for each value of \(i\):
05

i = 2

P(sum = 2) = 1/36, P(first die = 6 and sum = 2) = 0 \(P(first\_die\_6 | sum\_2) = \frac{0}{1/36} = 0\)
06

i = 3

P(sum = 3) = 2/36, P(first die = 6 and sum = 3) = 0 \(P(first\_die\_6 | sum\_3) = \frac{0}{2/36} = 0\)
07

i = 4

P(sum = 4) = 3/36, P(first die = 6 and sum = 4) = 0 \(P(first\_die\_6 | sum\_4) = \frac{0}{3/36} = 0\)
08

i = 5

P(sum = 5) = 4/36, P(first die = 6 and sum = 5) = 0 \(P(first\_die\_6 | sum\_5) = \frac{0}{4/36} = 0\)
09

i = 6

P(sum = 6) = 5/36, P(first die = 6 and sum = 6) = 1/36 \(P(first\_die\_6 | sum\_6) = \frac{1/36}{5/36} = \frac{1}{5}\)
10

i = 7

P(sum = 7) = 6/36, P(first die = 6 and sum = 7) = 1/36 \(P(first\_die\_6 | sum\_7) = \frac{1/36}{6/36} = \frac{1}{6}\)
11

i = 8

P(sum = 8) = 5/36, P(first die = 6 and sum = 8) = 1/36 \(P(first\_die\_6 | sum\_8) = \frac{1/36}{5/36} = \frac{1}{5}\)
12

i = 9

P(sum = 9) = 4/36, P(first die = 6 and sum = 9) = 1/36 \(P(first\_die\_6 | sum\_9) = \frac{1/36}{4/36} = \frac{1}{4}\)
13

i = 10

P(sum = 10) = 3/36, P(first die = 6 and sum = 10) = 1/36 \(P(first\_die\_6 | sum\_10) = \frac{1/36}{3/36} = \frac{1}{3}\)
14

i = 11

P(sum = 11) = 2/36, P(first die = 6 and sum = 11) = 1/36 \(P(first\_die\_6 | sum\_11) = \frac{1/36}{2/36} = \frac{1}{2}\)
15

i = 12

P(sum = 12) = 1/36, P(first die = 6 and sum = 12) = 1/36 \(P(first\_die\_6 | sum\_12) = \frac{1/36}{1/36} = 1\) So, the conditional probabilities for each value of \(i\) are as follows: \(P(first\_die\_6 | sum\_2) = 0\) \(P(first\_die\_6 | sum\_3) = 0\) \(P(first\_die\_6 | sum\_4) = 0\) \(P(first\_die\_6 | sum\_5) = 0\) \(P(first\_die\_6 | sum\_6) = \frac{1}{5}\) \(P(first\_die\_6 | sum\_7) = \frac{1}{6}\) \(P(first\_die\_6 | sum\_8) = \frac{1}{5}\) \(P(first\_die\_6 | sum\_9) = \frac{1}{4}\) \(P(first\_die\_6 | sum\_10) = \frac{1}{3}\) \(P(first\_die\_6 | sum\_11) = \frac{1}{2}\) \(P(first\_die\_6 | sum\_12) = 1\)

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Key Concepts

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

Probability Theory
Probability theory is the mathematical study of randomness and uncertainty. It provides a framework for quantifying how likely events are to occur within a specific context. At its foundation, probability theory uses a set of rules to calculate the likelihood, or probability, of various outcomes given a set of conditions or previous events.

For example, when rolling two fair dice, we can use probability theory to determine the likelihood of the first die landing on a 6. Probability is often expressed as a number between 0 and 1, where 0 indicates the event cannot happen, and 1 indicates it certainly will happen. Understanding probability theory is essential for a range of activities, from gambling and game strategies to making predictions in finance, meteorology, and many scientific disciplines.
Sample Space
In probability, the sample space is the set of all possible outcomes of a particular experiment. For the act of rolling two dice, the sample space consists of 36 possible results (6 outcomes from the first die multiplied by 6 outcomes from the second die). Each possible result is represented by a pair of numbers, each number reflecting the outcome of one die.

Understanding the sample space is crucial because it lays the groundwork for calculating probabilities. To accurately determine the probability of any event, you must be mindful of the entire set of possible outcomes, as this influences the chances of the event occurring. If we fail to account for all the possibilities, our probability calculations could be off, leading to incorrect conclusions about how likely or unlikely certain events are.
Probability Formula
The probability formula is a fundamental equation used to compute the likelihood of an event occurring. It is described as the number of favorable outcomes divided by the total number of possible outcomes in the sample space. In mathematical terms, the probability of an event A is denoted as \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

Conditional probability, an extension of this core idea, considers how the probability of event A changes if we know event B has occurred. The conditional probability formula is \( P(A | B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A | B) \) is the probability of A given B, \( P(A \cap B) \) is the joint probability of both A and B, and \( P(B) \) is the probability of B alone. This formula is particularly useful in situations where probabilities are affected by prior events, which is common in many practical applications like risk assessment and statistical inference.

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Most popular questions from this chapter

The probability of getting a head on a single toss of a coin is \(p\). Consider that \(A\) starts and continues to flip the coin until a tail shows up, at which point \(B\) starts flipping. Then \(B\) continues to flip until a tail comes up, at which point \(A\) takes over, and so on. Let \(P_{n, m}\) denote the probability that \(A\) accumulates a total of \(n\) heads before \(B\) accumulates \(m\). Show that $$ P_{n, m}=p P_{n-1, m}+(1-p)\left(1-P_{m, n}\right) $$

Fifty-two percent of the students at a certain college are females. Five percent of the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that (a) this student is female, given that the student is majoring in computer science; (b) this student is majoring in computer science, given that the student is female.

If \(0 \leq a_{i} \leq 1, i=1,2, \ldots\), show that $$ \sum_{i=1}^{x}\left[a_{i} \prod_{j=1}^{i-1}\left(1-a_{j}\right)\right]+\prod_{i=1}^{x}\left(1-a_{i}\right)=1 $$ HINT: Suppose that an infinite number of coins are to be flipped. Let \(a_{i}\) be the probability that the ith coin lands heads, and consider when the first head occurs.

There are 3 coins in a box. One is a two-headed coin; another is a fair coin; and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

Rank the following from most likely to least likely to occur. 1\. A fair coin lands on heads. 2\. Three independent trials, each of which is a success with probability .8, all result in successes. 3\. Seven independent trials, each of which is a success with probability .9, all results in successes.

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