probability help!!!!!! (3 Viewers)

[lookoutastroboy]

There is a spinner. It has 12 slots with 4 1s, 2 2s, 3 3s, one 4, one 5 and one 6. What is the probability of at least two different scores?



This is an elementary question from Yr 10 but i figured people from this forum would have little trouble to tackle it. If you can, please explain it in detail in how you got the answer =)










Thanks, lookoutastroboy

 

[lolokay]

'at least two different scores' - what do you mean by that? the question doesn't sound complete

 

[-tal-]

'at least two different scores' - what do you mean by that? the question doesn't sound complete

agreed. question is awfully ambiguous.

 

[lookoutastroboy]

here i've tried to take a screenshot of it (i hope u can see it - anywayz this question is from mathscape yr 10 extension text book)

thanks for the responses.
good luck solving it, lookoutastroboy

 

[lookoutastroboy]

here is a screenshot of the question (i hope u can see it guyz and galz)

thanks for the responses.
good luck solving it (this comes from mathsscape yr 10 ext text book btw).

 

[lookoutastroboy]

sorry for double post. thought 1st one didnt come through

 

[Trebla]

You should have said it was spun 3 times...lol

P(at least two different scores) = 1 - P(all same scores)

P(all the same scores) = P(three 1s) + P(three 2s) + .... + P(three 6s)

P(three 1s) = (4/12)³ = 1/27
P(three 2s) = (2/12)³ = 1/216
P(three 3s) = (3/12)³ = 1/64
P(three 4s) = (1/12)³ = 1/1728
P(three 5s) = (1/12)³ = 1/1728
P(three 6s) = (1/12)³ = 1/1728

P(all the same scores) = 1/27 + 1/216 + 1/64 + 1/1728 + 1/1728 + 1/1728 = 17/288

.: P(at least two different scores) = 1 - 17/288
= 271/288

 

[lookoutastroboy]

hey thanks to evryone who responded and tried to solve this question. Sorry for not putting in all of the question (my fault to many who could have solved it otherwise).
thanks trebla for solving !!!!!

 

[lookoutastroboy]

just one more question if you please?

11 Rachael purchases a townhouse ‘off the plan’ and the builder offers her a choice of colours
for interior painting. The walls can be cream, light blue or light green, and the ceilings vivid
white or off-white. Different colours may be chosen for different rooms but only two
colours can be used in any room (one for the ceiling and one for the walls). The townhouse
consists of four rooms; living room, bedroom, kitchen and bathroom. It is all so confusing
that Rachael makes random selections for each room.
a What is the number of possible colour schemes for Rachael’s townhouse?
b What is the probability that:
i the bedroom has cream walls and a vivid white ceiling?
ii the living room and the bedroom have the same colour schemes?
iii no ceiling is off-white?
iv all the walls are of the same colour?
v at least two rooms have identical colour schemes?


thanks again for all your help, lookoutastroboy (p.s. why is probability so hard!?)

 

[kurt.physics]

just one more question if you please?

11 Rachael purchases a townhouse ‘off the plan’ and the builder offers her a choice of colours
for interior painting. The walls can be cream, light blue or light green, and the ceilings vivid
white or off-white. Different colours may be chosen for different rooms but only two
colours can be used in any room (one for the ceiling and one for the walls). The townhouse
consists of four rooms; living room, bedroom, kitchen and bathroom. It is all so confusing
that Rachael makes random selections for each room.
a What is the number of possible colour schemes for Rachael’s townhouse?
b What is the probability that:
i the bedroom has cream walls and a vivid white ceiling?
ii the living room and the bedroom have the same colour schemes?
iii no ceiling is off-white?
iv all the walls are of the same colour?
v at least two rooms have identical colour schemes?


thanks again for all your help, lookoutastroboy (p.s. why is probability so hard!?)

a) This is a counting problem. There are 3 colours for the walls and 2 colours for the ceiling. Further more, there can only be two colours in a room. Seeing as both the wall and ceiling have to be painted. So for each room, there will be 3 x 2 = 6 ways to arrange these colours. This is because you can use any of the three wall colours and then from that choose anyone of the two ceiling colours, using the multiplication principal, this becomes 6 ways.

The multiplication principal is a basic principal in combinatorics, another, harder, way is to use a tree diagram, then add all the last branches.

Anyway, to continue, we have 6 ways to choose the colours for a room. There are a total of 8 rooms, and so, again using the multiplication principal (dont worry if you dont know), there is 6 x 8 ways or 48 ways.

b) Okay, we can draw a tree diagram to answer the question, the first branches are cream light blue light green

from these three branches there are the two branches vivid white vivid off white. The probability for the colour of each wall is 1/3 and for each ceiling it is 1/2. Then we follow the cream walls branch and the vivid white branch, multiply the probability and get 1/3 x 1/2 = 1/6

(We could also have realised that combination {cream, white} is one possible combination of the six, ergo the prob is 1/6)

ii) This is a hard one to explain, i would assume that the text book has some kind of formula for this but because i havent done probability in one and a half years i like to use tree diagrams.

For me, i visulize a tree diagram such that we have the three branches for the first rooms first wall colour and then extending from then the first room ceiling colour and then from those the second rooms wall colour and so on. As you could imagine, it looks messy.

P(b,w,b,w) = 1/3 x 1/2 x 1/3 x 1/2
= 1/6 x 1/6
= 1/36

This is if both the two rooms in blue and white. So if we imagine that we choose say g, w. Then the prob will be 1/36. So we postulate that we multiply the number of same colour combinations by this probability. I think there will be six that are the same, because we have 6 different ways of doing the colours.

So the probability is,

6 x 1/36 = 1/6

I will try do the other later, just going out right now

 

[kurt.physics]

Things changed, i'll continue

iii) P(off white) = (1/2)⁸
= 1/256
P(not off white) = 1 - 1/256
= 255/256

This is for all of the rooms, see the probability of it being not off white for one room is 1/2, but for two rooms it is 1/4 ie P(not, not) and so on. I dont think we have to consider the walls as it doesnt affect the ceiling.

And this kind of makes sense, you could imagine that it would be highly unlikely that every colour is the same only if you make it that way, but its random.

iv) This question is quite similar to the previous. The probability of, say, all of them being green is

(1/3)⁸ = 1/6561

and for cream it is

(1/3)⁸ = 1/6561

and blue

(1/3)⁸ = 1/6561

and so the probability of the rooms walls being the same colour is

3 x 1/6561 = 1/2187

I cannot help you with v as that part has escaped my memory.

Any way, there you go. I think thats right, does the textbook agree.

Cheers,
-Kurt

 

[lookoutastroboy]

oy kurt, thanks a lot for the help. It is just unfortunate that some of your answers are wrong and here are the solutions. By the way, at the start why did you say that there were 8 rooms when there were only 4?

a. 1296
b
i. right
ii. right
iii.1/16
iv. 1/27
v. 13/18






thanks so much !!!!!

 

[kurt.physics]

oy kurt, thanks a lot for the help. It is just unfortunate that some of your answers are wrong and here are the solutions. By the way, at the start why did you say that there were 8 rooms when there were only 4?

a. 1296
b
i. right
ii. right
iii.1/16
iv. 1/27
v. 13/18

thanks so much !!!!!

Sorry about that

Was the question saying that there are 4 rooms, where each room is the bath...

I thought the question was saying there were 4 rooms and bathroom etc

 

[lookoutastroboy]

no problems kurt, would u mind if u solved the ones u got wrong?
would be a great help =))) thanks in advance

 

[kurt.physics]

so for question b iii) the method there was wrong. Because its between off white and what ever the other white. If every ceiling was not off white, then it would be the other white. And because there are 4 rooms that means the probability

P(other white) = (1/2)⁴

which turns out to be 1/16

And for iv) working with 4 rooms we get

Prob. = 3 x (1/3)⁴

which turns out to be 1/27

 

[lookoutastroboy]

oy kurt, one thing man, my tree diagram in total has 24 options, do u think this is right? i got the 4 type of rooms in order and then the 3 wall colours to each of them with two of the ceiling colours sprouting out of that ,forming 24? but i dunno how a is 1296?

 

[kurt.physics]

With a, i would like to apologize about my ridiculous solution to it using combinatorics, see i started it yesterday and thought i had the go of it (looks like i will have some math to occupy myself.)

Anyway, the real solution is

3 x 2 x 3 x 2 x 3 x 2 x 3 x 2

This is 1296.

It seams we can use the multiplication principal, but it seams that i had used it wrong. We look at the first room, we can choose any 3 colours for the wall and 2 colours for the ceiling. So there are 3 x 2 or 6 ways. We can think about it as a tree diagram, from the three branches we put another 2 branches on each branch, so there will be 3 lots of 2 or 6.

So if you can imagine putting another three branches on to each of the 6 ceiling branches (the reason why we do this is because we are looking at the second room and we start again with the walls.)

Then there will be 6 x 3 branches, or 18. We just keep doing this, and you get the above bold and underlined diagram above.

I sincerely apoligise again.

 

[lookoutastroboy]

oy kurt, i had a tree diagram with the 1st room (living room) and then the 3 wall colours sprouting out of that with the ceiling colours sprouted out of each wall colour. Then i did the same for each room underneath it. I still dont understand the part bout using 18 branches from the ceiling colours because each room is separate from each other?

 

[lookoutastroboy]

would anyone be able to do the last question (at least two rooms have identical colour schemes - the probability of it) help would be much appreciated!!!

 

[Timothy.Siu]

okay

v)theres 6 different combinations...

so, say one of the rooms has a certain combination, the chance of the next room having a DIFFERENT colour scheme is 5/6, the next, would be 4/6, and the next 3/6 since one more is taken up each time
probability of all different is 6/6x5/6x4/6x3/6=5/18
so, 1- (6/6x5/6x4/6x3/6)=13/18.