Question

When two six-sided dice are rolled, there are 36 possible outcomes.

a. Find the probability that the sum is not 4. Express your first answer as a fraction in simplest form, and round your percent answer to the nearest whole percent.

b. Find the probability that the sum is greater than 5. Express your first answer as a fraction in simplest form, and round your percent answer to the nearest whole percent.

Answer 1

A. 33/36 or 92% B. 26/36 or 72%

A. List the possibilities that do have a sum of 4. (1,3) (3,1) (2,2).  3/36. subtract 3/36 from the whole to get probability it does not have a sum of 4.  36/36-3/36=33/36 or 92%

B. List the possibilities if is less than or equal to 5 because it will be a smaller amount than if it is greater and subtract like we did in the previous problem. (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). 10 total  36/36-10/36= 26/36 or 72%

Answer 2

Using the probability concept, it is found that there is a:

a) $\frac{33}{36} = 92\%$ probability that the sum is not 4.

b) $\frac{30}{36} = 83\%$ probability that the sum is greater than 5.

A probability is the number of desired outcomes divided by the number of possible outcomes.

Item a:

• There are 3 outcomes in which the sum is 4: (1,3), (2,2) and (3,1).
• Thus, there are 36 - 3 = 33 outcomes in which the sum is not 4.

The probability is:

$p = \frac{D}{T} = \frac{33}{36} = 92\%$

$\frac{33}{36} = 92\%$ probability that the sum is not 4.

Item b:

• There are 6 outcomes in which the 4 or less: (1,1), (1,2), (1,3), (2,1), (2,2) and (3,1).

• Thus, there are 36 - 6 = 30 outcomes in which the sum is greater than 4.

The probability is:

$p = \frac{D}{T} = \frac{30}{36} = 83\%$

$\frac{30}{36} = 83\%$ probability that the sum is greater than 5.

A similar problem is given at https://brainly.com/question/4818951