If a standard six-sided die is rolled twice, what is the probability that the second number rolled is not less than the first number rolled? Express your answer as a common fraction.

Step-by-step explanation:

Probabilities

When rolling a dice twice, we can get 36 different combinations of the numbers, starting from (1,1),(1,2),(1,3),... up to (6,4),(6,5),(6,6).

From those combinations, only those where the second number rolled is not less than the first number are favorable outcomes.

Thus, the favorable outcomes are:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),

(4,4),(4,5),(4,6),(5,5),(5,6),(6,6)}

The total number of favorable outcomes is: 21. Thus the required probability is:

p=\frac{21}{36}=\frac{7}{12}

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are