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Chapter 9: Problem 29

Find the probability for the experiment of tossing a six-sided die twice. The sum is odd and no more than 7.

Short Answer

Expert verified
The probability is \( \frac{12}{36} = \frac{1}{3}\).

Step by step solution

01

Identify the favorable outcomes

With two dice, to get an odd sum, one die must land on an even number and the other on an odd number. The even outcomes are [2,4,6] and the odd outcomes are [1,3,5]. Now let's proceed to sum up these possible pairs to identify those sums that are odd and less or equal to 7. The pairs matching these conditions are: (2,1), (2,3), (2,5), (4,1), (4,3), (6,1). Remember that order matters, so each of these counts as 2 (e.g. (1,2) and (2,1) are considered separate outcomes). Thus, there are a total of 12 favorable outcomes.
02

Calculate the probability

To calculate the probability, divide the number of favorable outcomes by the total number of outcomes. So, probability = \( \frac{favorable \ outcomes}{total \ outcomes} = \frac{12}{36}\)

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