Question

If you roll two number cubes and toss a coin at the same time, what is the possibility that the cubes will both show 6 and the coin will land HEADS up?

Answer 1

these are independent events

Answer 2

I need the possibility @Zarkon

Answer 3

if you roll one number cube, what is the probability of rolling a 6?

Answer 4

@WaVVessux
I'm sure you do. I'm not in the habit though of just giving out answers

Answer 5

Okay, we'll take an even smaller step. If you toss a coin, what is the probability it will land HEADS up?

Answer 6

@Zarkon that's why you explain it. Nobody asked for the answer, why would you even answer? I know they're independent events

Answer 7

@whpalmer4 50/50

Answer 8

okay, that means 0.5 is the probability of that event. Now, what is the probability of rolling 1 number cube and getting a 6?

Answer 9

@whpalmer4 1/6

Answer 10

right. P = 1/6

Now, we have independent events, so we multiply the probabilities to get the probability of all 3 things happening together.

Answer 11

Looking at it a different way, there are 72 different possible outcomes:

1 1 H
1 1 T
1 2 H
1 2 T
....
6 6 H
6 6 T
Only one of those outcomes is the desired 6 6 H, so the probability is ...

Answer 12

I a little shocked that you would know that they are independent but are unable to compute the probability.

Answer 13

@Zarkon and i'm a little shocked you're an ambassador and just use it to boss people around and be condescending. Then again, you probably get some sick kick out of bossing people around because you can't in real life

Answer 14

Sorry @whpalmer4 gimmie a sec

Answer 15

no problem, I'm making popcorn to eat during the fireworks :-)

Answer 16

so it's 1/6*3/1 right?

Answer 17

@whpalmer4

Answer 18

How do you figure that? you have 3 independent events, with probabilities 1/6, 1/6, and 1/2. With independent events, you multiply the probabilities to get the overall probability.

Alternative approach, enumerate all possible outcomes. Count successful outcomes, and divide by number of possible outcomes.

Same answer either way.

Answer 19

1/12? @whpalmer4

Answer 20

@WaVVessux
I did not insult you...therefore don't insult me. Also I fail to see how I "bossed you around"

Also I am not an ambassador.

Answer 21

3 probabilities: 1/6, 1/6, 1/2 multiply them together.

Answer 22

You're right, I realized just now you're a moderator. So therefore, you would think you would be a little more polite. I didn't insult you either, I called you no names. You are very rude and condescending. I've never talked to you before and given you a reason to be rude.I'm reporting you to @Preetha @Zarkon

Answer 23

@whpalmer4 sorry I just missed one 1/6

Answer 24

"i'm a little shocked you're an ambassador and just use it to boss people around and be condescending. Then again, you probably get some sick kick out of bossing people around because you can't in real life"

is an insult.

in particular... "Then again, you probably get some sick kick out of bossing people around because you can't in real life"

Answer 25

I've seen Zarkon interact with quite a few people (including you), and I wouldn't describe his behavior in any of those cases as rude and condescending. Just saying...

Answer 26

@Zarkon well,since it was a retort definitely. But i'm not just going to let you be rude. Don't be condescending and think that i'm just gonna take it because you're a mod.

Answer 27

this ... what i said..."I a little shocked that you would know that they are independent but are unable to compute the probability."

is an observation from teaching at the college level for over a decade.


for a problem like this if a student couldn't get the answer it is usually because they don't realize the events are independent or that they needed to multiply.

Answer 28

@whpalmer4 well,I've never talked or seen him interact with anyone. Its good, he talks to others nicely. But he was condescending to me. @Zarkon you don't know my details though, isn't this what people come here for? Help? All I wanted was an explanation, this wans't the first place I went to for help. It's not your right to judge whether I can or cannot get a simple problem. I struggle in math, and I get a lot of help. Not just from here either.

Answer 29

@Zarkon and even then sometimes I won't get it with all my help.

Answer 30

Let me ask you this: the problem statement clearly tells anyone who reads it that you need the probability. Why then did you say what you did, specifically to @Zarkon?

Answer 31

I didn't know the relationship between probability and independence.

Answer 32

thought they were two different things. Also I have 1/72 now as my answer

Answer 33

I was giving a hint....if you knew that hint then you could have said in the very beginning that you knew that they were independent instead of saying that you want probabilities (because clearly that is ultimately what you wanted.) from my perspective you sounded like someone that just wanted answers.

Answer 34

@Zarkon like I said, I thought independence and probability are two different things. I didn't know it had anything to do with the problem. Besides I hardly get answers here, if I do it's with explanation. And this place has a lot of rude users, so from my point of view you were just being sarcastic or something

Answer 35

oh and that's just how the problem came to. Said probability not independence. and 1/72 should be my answer right @whpalmer4 ?

Answer 36

For future reference...I am never sarcastic. I am about as direct as they come. So if you read something that I write and you think it is sarcasm...then you need to reread it.

For my part...I will not assume you are just asking for answers in any future post you make.

Answer 37

Thanks, and I will not assume you are being sarcastic in your future post as well. :) We cool?

Answer 38

Sure

Answer 39

@whpalmer4 um 1/72 I got as my answer?

Answer 40

I told you two different ways to compute the answer. Do it the other way to check your result...

Answer 41

okay sorry didn't see that

Answer 42

so you don't have to hunt:


Looking at it a different way, there are 72 different possible outcomes:

1 1 H
1 1 T
1 2 H
1 2 T
....
6 6 H
6 6 T
Only one of those outcomes is the desired 6 6 H, so the probability is ...

=========

Alternative approach, enumerate all possible outcomes. Count successful outcomes, and divide by number of possible outcomes.

Same answer either way.