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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

Show that $$ P\left(\bigcup_{i=1}^{n} E_{i}\right) \leqslant \sum_{i=1}^{n} P\left(E_{i}\right) $$ This is known as Boole's inequality. Hint: Either use Equation (1.2) and mathematical induction, or else show that \(\bigcup_{i=1}^{n} E_{i}=\bigcup_{i=1}^{n} F_{i}\), where \(F_{1}=E_{1}, F_{i}=E_{i} \bigcap_{j=1}^{i-1} E_{j}^{c}\), and use property (iii) of a probability.

For a fixed event \(B\), show that the collection \(P(A \mid B)\), defined for all events \(A\), satisfies the three conditions for a probability. Conclude from this that $$ P(A \mid B)=P(A \mid B C) P(C \mid B)+P\left(A \mid B C^{c}\right) P\left(C^{c} \mid B\right) $$
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