Question

You roll two standard number cubes. What is the probability that the sum is odd, given than one of the number cubes shows a 1? Show your work.

Answer 1

This question involves more complex probabilities. Whenever you see a question saying you roll 2 standard number cubes, always know your denominator will be 36. The reason for that is shown below: 

$\frac{x}{6} * \frac{x}{6}$

X is your favorable outcomes, and to find your probabilities, you multiply the two. Don't consider x as a variable. Im just giving you this equation because it is what you can set up the second you read the first sentence of a question.

Anyway, now I can see, the question asks, what is the probability that the sum is odd, given that one of the number cubes shows a 1. So we can plug in 1 for x, because that is the only value we can use. Now we have 1/6 * x/6. So the sum has to be odd, if I rolled a 1 on the second die, that wouldn't work, if I rolled a 2 on the second die, it works because 2 + 1 equals 3, if I rolled a 3 on the second die it doesn't work, if I rolled a 4 on the second die it works, if I rolled a 5 on the second die it doesn't work, and if I rolled a 6 on the second die it works. Out of the 6 possibilities, 3 of my numbers work, so the second x would equal 3. Therefore your equation is:

$\frac{1}{6} * \frac{3}{6} = \frac{3}{36}$

This simplifies into 1/12, which is equal to 0.08333 repeating. Therefore the probability of the sum being odd given that one of the number cubes shows a 1 is 0.08333333333 as a decimal or 8.333333333% as a percent. 


Hope this helps. Please rate, leave a thanks, and mark a brainliest answer. (Not necessarily mine). Thanks, it really helps! :D