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Chapter 1: Problem 11

If two fair dice are tossed, what is the probability that the sum is \(i, i=\) \(2,3, \ldots, 12 ?\)

Short Answer

Expert verified
The probabilities for the sum $i$ when two fair dice are tossed are as follows: \( P(i=2) = \frac{1}{36} \) \( P(i=3) = \frac{2}{36} \) \( P(i=4) = \frac{3}{36} \) \( P(i=5) = \frac{4}{36} \) \( P(i=6) = \frac{5}{36} \) \( P(i=7) = \frac{6}{36} \) \( P(i=8) = \frac{5}{36} \) \( P(i=9) = \frac{4}{36} \) \( P(i=10) = \frac{3}{36} \) \( P(i=11) = \frac{2}{36} \) \( P(i=12) = \frac{1}{36} \)

Step by step solution

01

List down all the possible outcomes

We have two dice being rolled, meaning there are 6 faces on each die. In every roll, there are 6 possible outcomes for each die which results in 6 x 6 = 36 different combinations for the two dice. The combinations are: (1,1), (1,2), ..., (1,6) (2,1), (2,2), ..., (2,6) . . . (6,1), (6,2), ..., (6,6)
02

Find the number of combinations for each sum i

Now, for each i ranging from 2 to 12, we will find the number of combinations where the sum of the two dice rolls equals i. Doing this, we get: - i = 2: 1 combination, (1,1) - i = 3: 2 combinations, (1,2) and (2,1) - i = 4: 3 combinations, (1,3), (2,2) and (3,1) - i = 5: 4 combinations, (1,4), (2,3), (3,2) and (4,1) - i = 6: 5 combinations, (1,5), (2,4), (3,3), (4,2) and (5,1) - i = 7: 6 combinations, (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) - i = 8: 5 combinations, (2,6), (3,5), (4,4), (5,3) and (6,2) - i = 9: 4 combinations, (3,6), (4,5), (5,4) and (6,3) - i = 10: 3 combinations, (4,6), (5,5) and (6,4) - i = 11: 2 combinations, (5,6) and (6,5) - i = 12: 1 combination, (6,6)
03

Calculate the probability for each sum i

Finally, to find the probability of each sum i, we divide the number of combinations for that sum by the total number of possible outcomes (36). Thus, we get the following probabilities: - P(i=2) = 1/36 - P(i=3) = 2/36 - P(i=4) = 3/36 - P(i=5) = 4/36 - P(i=6) = 5/36 - P(i=7) = 6/36 - P(i=8) = 5/36 - P(i=9) = 4/36 - P(i=10) = 3/36 - P(i=11) = 2/36 - P(i=12) = 1/36 So, these are the probabilities of the sum i being 2, 3, ..., 12 for two fair dice being tossed.

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Most popular questions from this chapter

We say that \(E \subset F\) if every point in \(E\) is also in \(F\). Show that if \(E \subset F\), then $$ P(F)=P(E)+P\left(F E^{c}\right) \geqslant P(E) $$

A box contains three marbles: one red, one green, and one blue. Consider an experiment that consists of taking one marble from the box then replacing it in the box and drawing a second marble from the box. What is the sample space? If, at all times, each marble in the box is equally likely to be selected, what is the probability of each point in the sample space?

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellows were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\), since he would then be one of two prisoners. What do you think of the jailer's reasoning?

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?

A coin is to be tossed until a head appears twice in a row. What is the sample space for this experiment? If the coin is fair, what is the probability that it will be tossed exactly four times?
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