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mr_coffee

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Hello everyone, I'm having troubles grasping the answer to this question. I'm studying for the exam so the answers are given in full by the professor.

Here is the question:

Two faces of six sided die are painted red, two are painted blue , and two are painted yellow. The die is rolled 3 times, and the colors that appeared face up on the first, second and thrid rollsa re recorded.

Consider the situation described above:

a. Find the probabily of the event that

b. find the probabily of the event that at

a: Exactly one color being red can happen in 12 ways. (One can list them all and count them. Or one can use

the multiplication rule from the next section to calculate how many 3-letter sequences will show up in the list:

there are 3 positions in which the 'R' can occur, and the two remaining positions can be filled with B or Y in 2x2 ways. So there will be 3 x 2 x 2 = 12, 3 lettered sequences in the list.

So the probabilty is 12/(3x3x3) = 4/9 .

I'm confused on why there are 3 postions in which R can occur, is it because if you roll it 3 times, then that means its possible to get R, every single time you roll the die?

I'm trying to apply the multipcation rule, so i need to break this down into a step by step process in which its covering all the cases.

So would my first step be:

On roll 1, you get a red face

step #2.

On roll 2, you get either a blue or yellow

step #3.

On roll 3, you get either a blue or yellow

I don't see how this can be applied to the mulicpation rule becuase you could end up with more than 1 choice depending on what happened before hand, can someoen explain?

I can see how you can get how many possible total combinations in 3 rolls, like

step 1:

Roll the die, 3 things could pop up

step 2:

roll the die, 3 things could pop up

step 3:

roll the die, 3 things could pop up

so the total number of ways that Red, blue or yellow will appear is 3x3x3 = 27

b: At least one color being red can happen in 19 ways. (Again, one can list them all and count them, or one

can calculate that 2^3 = 8 of the 3^3 = 27 possible outcomes involve only blue and yellow, so the remaining

27 − 8 = 19 outcomes include at least one red.) Therefore, the probability of at least one red is 19/27, which

is about 70.4%.

So for the at least 1 case, can you always use this rule, find out 2^3, excluding the 1 your looking for in this case Red face comes up. Then find the total number of possible outcomes if all 3 where rolled which is 3^3, then subtract the 2 and you'll be left with how many times red will occur?Thanks any help would be great.

Thanks

Here is the question:

Two faces of six sided die are painted red, two are painted blue , and two are painted yellow. The die is rolled 3 times, and the colors that appeared face up on the first, second and thrid rollsa re recorded.

Consider the situation described above:

a. Find the probabily of the event that

**exactly one**of the colors that appears face up is red.b. find the probabily of the event that at

**least one**of the colors that appears face up is red.Answer:a: Exactly one color being red can happen in 12 ways. (One can list them all and count them. Or one can use

the multiplication rule from the next section to calculate how many 3-letter sequences will show up in the list:

there are 3 positions in which the 'R' can occur, and the two remaining positions can be filled with B or Y in 2x2 ways. So there will be 3 x 2 x 2 = 12, 3 lettered sequences in the list.

So the probabilty is 12/(3x3x3) = 4/9 .

I'm confused on why there are 3 postions in which R can occur, is it because if you roll it 3 times, then that means its possible to get R, every single time you roll the die?

I'm trying to apply the multipcation rule, so i need to break this down into a step by step process in which its covering all the cases.

So would my first step be:

On roll 1, you get a red face

step #2.

On roll 2, you get either a blue or yellow

step #3.

On roll 3, you get either a blue or yellow

I don't see how this can be applied to the mulicpation rule becuase you could end up with more than 1 choice depending on what happened before hand, can someoen explain?

I can see how you can get how many possible total combinations in 3 rolls, like

step 1:

Roll the die, 3 things could pop up

step 2:

roll the die, 3 things could pop up

step 3:

roll the die, 3 things could pop up

so the total number of ways that Red, blue or yellow will appear is 3x3x3 = 27

b: At least one color being red can happen in 19 ways. (Again, one can list them all and count them, or one

can calculate that 2^3 = 8 of the 3^3 = 27 possible outcomes involve only blue and yellow, so the remaining

27 − 8 = 19 outcomes include at least one red.) Therefore, the probability of at least one red is 19/27, which

is about 70.4%.

So for the at least 1 case, can you always use this rule, find out 2^3, excluding the 1 your looking for in this case Red face comes up. Then find the total number of possible outcomes if all 3 where rolled which is 3^3, then subtract the 2 and you'll be left with how many times red will occur?Thanks any help would be great.

**EDIT: a simple example of the multipcation rule that i understand is the following:**

Suppose a computer installation has 4 input/output units (a, b, c, and d) and 3 CPU's x,y, z. Any input/ouput can be paired with any central processing unit. How many ways are there to pair an input/ouptu unit with a central processing unit?

Step 1: Choose input/ouput unit.

step 2: choose the CPU

Well in step 1, there are 4 ways to choose the input/output unit, either a, b, c, or d

Step is there are 3 ways to choose the cpu, either x, y or z,

so the total number of ways is 4x3 = 12.

I get this but i don't see how the top problem can be transfored into this.Suppose a computer installation has 4 input/output units (a, b, c, and d) and 3 CPU's x,y, z. Any input/ouput can be paired with any central processing unit. How many ways are there to pair an input/ouptu unit with a central processing unit?

Step 1: Choose input/ouput unit.

step 2: choose the CPU

Well in step 1, there are 4 ways to choose the input/output unit, either a, b, c, or d

Step is there are 3 ways to choose the cpu, either x, y or z,

so the total number of ways is 4x3 = 12.

I get this but i don't see how the top problem can be transfored into this.

Thanks

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