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Chapter 1: Problem 31

What is the conditional probability that the first die is six given that the sum of the dice is seven?

Short Answer

Expert verified
The conditional probability that the first die is a six given that the sum of the dice is seven is \(\frac{1}{6}\).

Step by step solution

01

Determine the sample space for rolling two dice

When rolling two six-sided dice, there are a total of six possible outcomes for each die. This results in a sample space of size 6 x 6 = 36 possible outcomes (ordered pairs).
02

List the favourable outcomes for event A

Event A is the event that the first die shows a six. We can list these outcomes as follows: \((6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\). There are 6 favourable outcomes for event A.
03

List the favourable outcomes for event B

Event B is the event that the sum of the dice is seven. We can list these outcomes as follows: \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\). There are 6 favourable outcomes for event B.
04

Find the favourable outcomes for both A and B events

To find the favourable outcomes for both events A and B, we look for outcomes that are in both event A and event B. This results in the outcome \((6,1)\), which is the only outcome where the first die is a six and the sum of the dice is seven.
05

Calculate the conditional probability P(A|B)

Now we can apply the conditional probability formula: P(A|B) = P(A and B) / P(B). First, we find P(A and B), which is the probability of the favourable outcome \((6,1)\). Since there is only one favourable outcome and there are 36 possible outcomes in total, P(A and B) = 1/36. Next, we find P(B), which is the probability of the event B (the sum of the dice is seven). There are 6 favourable outcomes for event B and 36 possible outcomes in total, so P(B) = 6/36 = 1/6. Finally, we calculate P(A|B) = P(A and B) / P(B) = (1/36) / (1/6) = (1/36) * (6/1) = 1/6. Thus, the conditional probability that the first die is a six given that the sum of the dice is seven is 1/6.

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Most popular questions from this chapter

Urn 1 contains two white balls and one black ball, while urn 2 contains one white ball and five black balls. One ball is drawn at random from urn 1 and placed in urn 2. A ball is then drawn from urn 2. It happens to be white. What is the probability that the transferred ball was white?

Two cards are randomly selected from a deck of 52 playing cards. (a) What is the probability they constitute a pair (that is, that they are of the same denomination)? (b) What is the conditional probability they constitute a pair given that they are of different suits?

Suppose all \(n\) men at a party throw their hats in the center of the room. Each man then randomly selects a hat. Show that the probability that none of the \(n\) men selects his own hat is $$ \frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-+\cdots \frac{(-1)^{n}}{n !} $$ Note that as \(n \rightarrow \infty\) this converges to \(e^{-1}\). Is this surprising?

Suppose that 5 percent of men and \(0.25\) percent of women are color-blind. A color-blind person is chosen at random. What is the probability of this person being male? Assume that there are an equal number of males and females.

Consider two boxes, one containing one black and one white marble, the other, two black and one white marble. A box is selected at random and a marble is drawn at random from the selected box. What is the probability that the marble is black?
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