Question

you roll a number cube twice. find P(even, then not 2). write the probability as a fraction in simplest form

Answer 1

P(even number):
3/6 or 1/2
P( not 2):
5/6

5/6×1/2=5/12 is your final answer. Hope it help!

Answer 2

Using the concept of probability, when a number dice rolled twice then P(even, then not 2) is: 5/12

What is probability?

"Probability is a branch of mathematics which deals with finding out the likelihood of the occurrence of an event."

Formula of the probability of an event A is:

$P(A)=\frac{n(A)}{n(S)}$

where, $n(A)$ is the number of favorable outcomes

$n(S)$ is the total number of events in the sample space.

From given situation,

Consider an experiment of rolling a number dice twice

Let, event A: getting even number

and event B: not getting 2

Sample space, S = {1, 2, 3, 4, 5, 6}

⇒ n(S) = 6

Since A = {2, 4, 6}, n(A) = 3

⇒ $P(A)=\frac{3}{6}$

⇒ $P(A)=\frac{1}{2}$

For event B, B = {1, 3, 4, 5, 6}

⇒ $n(B)=5$

⇒ $P(B)=\frac{5}{6}$

Now, P(even, then not 2) is,

$=P(A)\times P(B)$

$=\frac{1}{2} \times \frac{5}{6} }$

$=\frac{5}{12}$

Hence, P(even, then not 2) = 5/12

Learn more about probability here:

https://brainly.com/question/11234923

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