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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

Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining \(p\), the probability that each hand has an ace. Let \(E_{i}\) be the event that the \(i\) th hand has exactly one ace. Determine \(p=P\left(E_{1} E_{2} E_{3} E_{4}\right)\) by using the multiplication rule.

In Laplace's rule of succession, suppose that the first \(n\) flips resulted in \(r\) heads and \(n-r\) tails. Show that the probability that the \((n+1)\) st flip turns up heads is \((r+1) /(n+2)\). To do so, you will have to prove and use the identity $$ \int_{0}^{1} y^{n}(1-y)^{m} d y=\frac{n ! m !}{(n+m+1) !} $$

Suppose that we want to generate the outcome of the flip of a fair coin but that all we have at our disposal is a biased coin which lands on heads with some unknown probability \(p\) that need not be equal to \(\frac{1}{2}\). Consider the following procedure for accomplishing our task. 1\. Flip the coin. 2\. Flip the coin again. 3\. If both fli?s land heads or both land tails, retum to step 1 . 4\. Let the result of the last flip be the result of the experiment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

Each of 2 cabinets identical in appearance has 2 drawers. Cabinet \(A\) contains a silver coin in each drawer, and cabinet \(B\) contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

There are two local factories that produce radios. Each radio produced at factory \(A\) is defective with probability \(.05\), whereas each one produced at factory \(B\) is defective with probability.01. Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory \(A\) or factory \(B\). If the first radio that you check is defective, what is the conditional probability that the other one is also defective?

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