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Chapter 9: Problem 30

Find the probability for the experiment of tossing a six-sided die twice. The sum is odd or prime.

Short Answer

Expert verified
The probability for the sum being odd or prime when a die is tossed twice is \(\frac{3}{4}\).

Step by step solution

01

Determining the Sample Space

Since a die has six faces, there are six possible outcomes each time it is thrown. So, for two dice, the total number of possible outcomes (sample space) is \(6 \times 6 = 36\).
02

Find the count for odd sum outcomes

Odd sum is possible when one of the dice results in an even number (either 2, 4 or 6) and the other results in an odd number (either 1, 3 or 5). So the possibilities are 3 ways for an even number times 3 ways for an odd number. That is \(3 \times 3 = 9\) odd sum possibilities. However, this can happen for both dice (either the first die being even and the second die being odd, or vice versa). So, total odd sums possible are \(9 + 9 = 18\).
03

Number of sums which result in a prime number

The prime sums that are possible (2, 3, 5, 7) can be gotten by: (1, 1) for 2, (1, 2) and (2, 1) for 3, (1, 4), (4, 1), (2, 3), (3, 2) for 5 and (3, 4), (4, 3), (5, 2), (2, 5), (1, 6), (6, 1) for 7. There are total 15 prime sum outcomes. Please note that (1, 2) and (2, 1) are two distinct outcomes.
04

Identify Overlapping Outcomes

On counting, you will realize that we have counted some sums twice in Step 2 and 3 because they fall into the categories of both odd sums and prime sums. For example, (1, 2), (3, 2), (1, 4), etc. are common to both categories of outcomes. We count the overlapping sums once. On listing down such cases, we get: (1, 2), (2, 1), (2, 3), (3, 2), (1, 4), and (4, 1), total 6 outcomes.
05

Calculating the Probability

The total number of favorable outcomes will be the sum of odd sum outcomes and prime sum outcomes, subtracting the overlapping outcomes. We have \(18+15-6 = 27\) favorable outcomes. Therefore, the probability of this event is the ratio of the number of favorable outcomes to the total outcomes, which is \(\frac{27}{36}\) or \(\frac{3}{4}\).

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Most popular questions from this chapter

Determine whether the statement is true or false. Justify your answer. Rolling a number less than 3 on a normal six-sided die has a probability of \(\frac{1}{3}\) . The complement of this event is to roll a number greater than \(3,\) and its probability is \(\frac{1}{2}\) .

Approximation In Exercises \(79-82,\) use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise \(79,\) use the expansion \((1.02)^{8}=(1+0.02)^{8}\) $$=1+8(0.02)+28(0.02)^{2}+\cdots+(0.02)^{8}$$ $$(1.02)^{8}$$

True or False? In Exercises 93 and \(94,\) determine whether the statement is true or false. Justify your answer. The Binomial Theorem could be used to produce each row of Pascal's Triangle.

Proof In Exercises \(99-102,\) prove the property for all integers \(r\) and \(n\) where \(0 \leq r \leq n .\) The sum of the numbers in the \(n\) th row of Pascal's Triangle is \(2^{n}\) .

Expanding a Complex Number In Exercises \(73-78\) , use the Binomial Theorem to expand the complex number. Simplify your result. $$(2-i)^{5}$$
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