You roll a number cube twice. Find P(even, then not 2). Write the probability as a fraction in
simplest form.
I don't understand this question at all plz help.
P(even) = 1/2
P(not 2) = 5/6
so just multiply them
Thanks
To find the probability of rolling an even number first, and then not rolling a 2 on a number cube, we need to understand a few things.
First, a number cube is a six-sided object, commonly known as a dice, with numbers 1 to 6 on each face.
Secondly, when we roll the number cube twice, it means that we roll it once and then roll it again consecutively.
Now, let's break down the problem step by step.
Step 1: Identify the possible outcomes for each roll.
When we roll a number cube, the possible outcomes are the numbers 1, 2, 3, 4, 5, and 6.
Step 2: Find the outcomes that fulfill the condition "even, then not 2".
To fulfill the condition, we need the first roll to be an even number and the second roll to be any number except for 2.
Looking at our possible outcomes, the even numbers are 2, 4, and 6. And for the second roll, we have 5 options since we need any number except for 2.
So, the outcomes that fulfill the condition are: (2, 1), (4, 1), (6, 1), (2, 3), (4, 3), (6, 3), (2, 4), (4, 4), (6, 4), (2, 5), (4, 5), (6, 5), (2, 6), (4, 6), (6, 6).
Step 3: Determine the total possible outcomes for both rolls.
Since each roll has 6 possible outcomes, the total number of outcomes for both rolls is 6 * 6 = 36.
Step 4: Calculate the probability by dividing the outcomes fulfilling the condition by the total possible outcomes.
The outcomes fulfilling the condition are 15 (as derived in Step 2).
So, the probability of rolling an even number first and then not rolling a 2 is 15/36.
To simplify the fraction, we can divide both the numerator and denominator by their greatest common divisor, which is 3.
15/36 can be simplified as 5/12.
Therefore, the probability is 5/12.