A person rolls two dice. What are the odds in favor of throwing at least an eight?
There are 6 ways to get a 7
there are 15 ways to get more than 7 (your case) and
there are 15 ways to get under 7
prob of at least an eight = 15/36 =5/12
To find the odds in favor of throwing at least an eight when rolling two dice, we first need to determine the number of favorable outcomes and the total number of possible outcomes.
We can start by listing all the possible outcomes when rolling two dice. Each die can have six possible outcomes (numbers 1 to 6), so the total number of outcomes is 6 * 6 = 36.
To determine the favorable outcomes, we need to find all the combinations that result in a sum of at least eight. Let's go through the possibilities:
- To get an eight, we have the following combinations: (2,6), (3,5), (4,4), (5,3), and (6,2). This gives us 5 favorable outcomes.
- To get a sum of nine, we have: (3,6), (4,5), (5,4), and (6,3). This gives us 4 favorable outcomes.
- To get a sum of ten, we have: (4,6), (5,5), and (6,4). This gives us 3 favorable outcomes.
- To get a sum of eleven, we have: (5,6) and (6,5). This gives us 2 favorable outcomes.
- Finally, to get a sum of twelve, we have: (6,6). This gives us 1 favorable outcome.
Adding up all the individual favorable outcomes, we have a total of 5 + 4 + 3 + 2 + 1 = 15 favorable outcomes.
Therefore, the odds in favor of throwing at least an eight when rolling two dice are 15 favorable outcomes out of a total of 36 possible outcomes.
To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 3. This gives us 15/36.
In its simplest form, the odds in favor of throwing at least an eight would be 5/12.