Probability of Getting Red at Least Once in Two Spins of a Wheel with 8 Colors

[pace]

uh, I mean probability with independent outcomes, heh.

Homework Statement

A wheel I'm spinning twice has 8 different colours(among them the colour red). It will then give two different outcomes. Q: What is the probability that the outcome will give the colour red at least once.

The Attempt at a Solution

((1/8)x(7/8))x2/3

or

((1/8)x(7/8))x((1/8)(1/8))

Neither gives the result 0.234, which is the answer. 0_o

 

[tiny-tim]

The Attempt at a Solution

((1/8)x(7/8))x2/3

or

(1/8)x(1/8)

Neither gives the result 0.234, which is the answer. 0_o

Hi pace!

(1/8)x(1/8) is the probability for both spins red.

(1/8)x(7/8) is the probability for first-spin-not-red & second-spin-red.

But where does your 2/3 come from?

 

[pace]

Hi there :)

Oh sorry, the last one I meant ((1/8)x(7/8))x((1/8)(1/8))

I'm thinking the 2/3 is the "red at least once"...

 

[tiny-tim]

umm … ((1/8)x(7/8))x((1/8)(1/8)) is the probability for first-spin-not-red & second-spin-red & thrid-spin-red & fourth-spin-red … but there's only two spins!

You're really confused, aren't you?

Let's see … make the question as simple as possible … let's write R for red, and N for not-red.

Then there's four possiblities: RR, RN, NR, and NN.

You're only interested in the total of the first three (ie, not NN).

So how do you count them?

 

[pace]

Ah, that gives four spins..! Yep, it's a mess in here :) I wrote down two answers that seemed the closest to what I felt would give the answer, eheh.

Um, ah yes, that gives 3/4, yes?

 

[pace]

hm, maybe I see, I start at of at the wrong place? I started thinking off at P(a)xP(b).

But...um.. so.. RR((1/8)x(1/8)) x RN(1/8)x(7/8) x NR(7/8)x(1/8) doesn't work ?

But then I miss Probable/Possible(NN being here)

(Work. I'm thinking I should maybe do more of the easier ones first)

 

[tiny-tim]

pace, why do you keep writing "x"?

(oh, you did that in your ((1/8)x(7/8))x((1/8)(1/8)) also, I think)

It's "+".

(btw, your answer 3/4 would have been right if it was a coin, so I think you've got the principle.

But you don't seem able to write it clearly in mathematics.)

Hint: 0.234 = 15/64 - does that help?

 

[pace]

Ok, no I really meant multiplied. I'll be back. Thanks a lot for help. Yeah, thinking like that helps me sometimes.

 

[pace]

7/8 + 8/8 ? The.. 7/8 would stand for no red, the 8/8 for all red, but that wouldn't make sense.. 7/8 + 4/8 + 4/8.. no.

Um.. arg, I don't get it.

 

[tiny-tim]

fractions!

7/8 + 8/8 ? The.. 7/8 would stand for no red, the 8/8 for all red, but that wouldn't make sense.. 7/8 + 4/8 + 4/8.. no.

Um.. arg, I don't get it.

Oh pace, you're no good at fractions, are you?

7/8 + 8/8 = 15/8.

Try again!

 

[pace]

hehe sorry. I don't know where to begin maybe. Well I have a B from 1mx actually, but it is a bit too long ago. I'm having a hard time only with probablity .

um, (1/8)x(8/8)+(7/8)(1/8) ?!.. no..

 

[pace]

That gives two results. I'm leaving out the 3/4(Probable/Possible(sp?!)) of course, but I don't seem to make that add up: Like (1/8)x(8/8)+(7/8)(1/8) (a third spin here? But that would make over 15). And then add four divisions at the bottom(Possible results), but that would makes a lot of counting, and it doesn't seem to me that that would be the way.

 

[tiny-tim]

um, (1/8)x(8/8)+(7/8)(1/8) ?!.. no..

… yes … !

ok, work backwards - you know it's (1/8)x(8/8)+(7/8)(1/8).

So, what is (1/8)x(8/8) the probability of?

And what is (7/8)x(1/8) the probability of?

 

[pace]

lol. I don't get the 8/8. .. ah it's 'no red' of course... right? *hits my own head* Why didn't I think that way... I got to think math to language/pictures ?! I'm afraid of probability. Cause I'm much better at maths than language and probablity seems to me to be more language oriented(?) blah blah blah.

(1/8)x(8/8) is NRxNN + (7/8)x(1/8) RRxRN.

But I was sure I had to bring in the Probable/Possible in there somewhere.. With a whole ( math ) / ( math )

 

[pace]

No, where's the RR?

 

[pace]

What's 7/8 ? (checking my books)

 

[tiny-tim]

(1/8)x(8/8) is NRxNN + (7/8)x(1/8) RRxRN.

No!

N is no-red. R is red.

So NR is first-not-red + second-red.

And NRxNN would be first-not-red + second-red + third-not-red + fourth-not-red, if you had 4 spins (and you wouldn't need to write the "x").

So … try again … what is (1/8)x(8/8)? … and then … what is (7/8)x(1/8)?

 

[pace]

checked.

"N is no-red. R is red.", yes, this I get... something like it.

You mean a x between second-red and third-not-red right?

at least one red x no red + (but shouldn't it stand 2/8 here, that would give probabllity for at least two reds) x at least one red. gaaawd, I'm so confused.

 

[tiny-tim]

at least one red x no red

No no no …

That doesn't even make any sense, does it?

(if there's at least one, how can there be none? )

Try again … begin a sentence "(1/8)x(8/8) is the probability of … "

 

[pace]

RNxNR:(R(at least one red))(N(no red))+(N(No red))(R(at least one red)) . But where's the RR and the other NR?

... (R(1/8))(N(8/8)+(N(7/8)x(R(1/8))... No..


or R(probablility of at least one red)N(probability of no red) + NR huh.



... Thinking :)

 

[tiny-tim]

(R(at least one red))(N(no red))+(N(No red))(R(at least one red)) . But where's the RR and the other NR?

Getting closer …

But 1/8 does not mean "at least one red", does it?

What does it mean?

(And our R does not mean "at least one red".
It just means "red".
If the R is on the left (as in RN), then it's first-spin-red.
If the R is on the right (as in RN), then it's second-spin-red.)

So what does 1/8 mean? … try again!

 

[pace]

(R(probability of at least one red)(1/8))(N(no red)(8/8))+(NN(?!)(7/8))x(R(probability of at least one red)(1/8))

 

[pace]

heh. um.. 1/8 means 1 divided by 8, lol. ? But I was sure previous in my book that meant the probability of at least one colour.

*really hungry, buys something to eat*

 

[pace]

It means 1 out of 8...

 

[tiny-tim]

heh. um.. 1/8 means 1 divided by 8, lol. ? But I was sure previous in my book that meant the probability of at least one colour.

No … it means the probability of exactly one colour! (in this case, red.)

All the basic probabilities you wil be dealing with are of exactly something.

You would have to add some of them to get "at least" something.

Now try again … (1/8)x(8/8) is the probability of …

*really hungry, buys something to eat*

Well, at least it's giving you an appetite! ​

 

[pace]

Yeah. I've noticed homework does that :)

probability of one red x probability of ... lol. 7/8 is no red(all the other colours)?, but that makes 8/8...? ... a big nothing? yaaarwg No, all the colors? But that's neither R or N.

 

[pace]

one red x all + no red x one red..

 

[pace]

oh. ah. Now I maybe see the / (?)

 

[pace]

all is the possible? and the rest is the probable(3). That would make perfect sense(?) uhm. It's as if you make a division into a sentence.

 

[tiny-tim]

probability of one red x probability of ... lol. 7/8 is no red(all the other colours)?,

Yes, that's right!

1/8 is probability of red;
and 7/8 is probability of not red.

so (7/8)(1/8) is probability of first-spin-not-red and second-spin-red.

Now what is (1/8)(8/8)?

And why is (1/8)(8/8) + (7/8)(1/8) the right answer?

 

[pace]

So.. you'd have to multiply the 'no red x one red' with 'all' to make it a whole division(sp?)

Why I'm really thinking this way I'm not totally sure.

 

[pace]

It's one red x all colours.

Why it's the right answer presents a difficulty :)

 

[pace]

But I'm thinking as above: " all is the possible and the rest is the probable(3). That would make perfect sense(?) uhm. It's as if you make a division into a sentence. you'd have to multiply the 'no red x one red' with 'all' to make it a whole division "

 

[pace]

(1/8)(8/8) + (7/8)(1/8)

Why is the 1(8/8) at just one side?.. Because it doesn't have to be more of it? I'm a little confused at it's presentation. How do I transform it from that, to ((1/8) + (7/8)(1/8)(probable)) / (8/8/(possible))

 

[pace]

But that woulnd't make sense above cause then 8/8 is multiplied into both factors..