# The probability of rolling a die and not getting a 2 or 4

## Answers

Besides 2 or 4, there are four other possibilities, of 6, that could occur: 1, 3, 5, 6.

Thus, the P of rolling a die and not getting a 2 or 4 is 4/6, or 2/3.

2/12 bc if you think of the probability of rolling a 4 on a dice once which is 1/6, you simply just multiply 1/6 × 1/6 to get the probability for 2 rolls. Hoped that helped.

.0000214

Step-by-step explanation:

i think its wrong but hope it helps

0.261

Step-by-step explanation:

We use the Bernoulli trial method to solve this. The form of the answer goes thus:

P = nCr * x^2 * y^6

Where n is the total number of trials which is 8, while r is the expected number of success, 2 in this case. x is the probability of success 1/6 while y is the probability of failure 5/6 in this case.

P= 8C2 * (1/6)^2 * (5/6)^6

P= 28 * 0.028 * 0.335 = 0.261

1/6

Step-by-step explanation:

there are 6 sides/numbers on a die. 1 roll with a possibility of 6 outcomes.

2/6 (33.33%)

Step-by-step explanation:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Because there are six possibilities and only one of those possibilities will yield a six, we can find the probability to be [tex]\frac{1}{6}[/tex]

C. 0.000021

Step-by-step explanation:

Each roll of the die is an independent event.

The probability of rolling a 1, then a 2, then a 3, then a 4, then a 5, then a 6 is the product of the 6 individual probabilities.

Probability of rolling a 1:

There are 6 possible outcomes, rolling 1, 2, 3, 4, 5, or 6.

There is one desired outcome, rolling a 1.

p(1) = 1/6

Probability of rolling a 2:

Similar to rolling a 1, there is 1 desired outcome,rolling a 2, out of 6 possible outcomes.

p(2) = 1/6

The same is true for rolling a 3, 4, 5, or 6. Each one has a probability of 1/6.

p(1, 2, 3, 4, 5, 6, in that order) = p(1) * p(2) * p(3) * p(4) * p(5) * p(6)

= (1/6)^6

= 1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6

= 1/46,656

= 0.000021433

C. 0.000021

it would be 2/3

Step-by-step explanation:

may be 5/2