I HATE probability! I have a question, too.. :D (1 Viewer)

[matty_boi17]

What a stupid, stupid topic!

Okay, so I'm doing the 2003 paper, and I'm doing question 27, b, i. The question is:

Two unbiased dice are thrown. The dice each have six faces. The faces are numbered 1, 2, 3, 4, 5 and 6. What is the probability that neither shows a six?

According to the marking notes, the way I did it was wrong... and I have absolutely no clue when it comes to questions like this. ANY help would be appreciated.

The next question is:

Dale plays a game with these dice. There is no entry fee. When the dice are thrown:

Dale WINS $20 if both show a 6.
Dale WINS $2 if there is only one 6.
He LOSES $2 if neither shows a 6.

Now, because the BOShitheads didn't put the answers up, I can't see if I'm correct. My working was:

(2/12x20)+(1/12x2)+(10/12x-2) = $1.83 financial expectation.

Did I do this right?

 

[Sickle]

Try something like:

5/6 x 5/6 = 25/36

I dunno I suck at probability... give me trig any day

 

[Lazarus]

P(neither 6) = P(not 6 on first die) * P(not 6 on second die)
             = 5/6 * 5/6
             = 25/36

 

[JayWalker]

What a stupid, stupid topic!
Dale plays a game with these dice. There is no entry fee. When the dice are thrown:

Dale WINS $20 if both show a 6.
Dale WINS $2 if there is only one 6.
He LOSES $2 if neither shows a 6.

Now, because the BOShitheads didn't put the answers up, I can't see if I'm correct. My working was:

(2/12x20)+(1/12x2)+(10/12x-2) = $1.83 financial expectation.

Did I do this right?

Your not actually asking a question. im assuming the question was, Because it is General maths, what ammount of money would Dale expect to walk out with..

Answer is here:

Probability he WINS 20 is 1/36 (1/6 * 1/6)
Probability he WINS 2 is 15/36 (1/6 * 5/6)
Probability he LOSES 2 is 25/36 (5/6 * 5/6)

Therefore, you multiply each by their money ammount..

$20 WIN:::::::::$2 WIN::::::::::$2 LOSE (Thats why its -2 not +2)
(20 * [1/36]) + (2 * [15/36]) + (-2 * 25/36)

= $0

Therefore he should expect to walk out as poor as he was when he started

Thats ASSUMING that i got the question right, could you post the real question?

(2/12x20)+(1/12x2)+(10/12x-2) = $1.83 financial expectation.

Nah you didnt do that right... youve just ADDED them not multiplied..

You got 2 / 12 by adding 1/6 + 1/6.. you have to multiply...

Probability(getting a 6) = 1/6
Probability(getting a 6 and then a 6) = 1/6 * 1/6

You will best understand by drawing a probability tree, then at the end of each branch writing what outcome it is, then count the number of favorable outcomes divided by the number of possible...

After all Probibility is (No. Preferable) / (No. Available)

 

[Jago]

The next question is:

Dale plays a game with these dice. There is no entry fee. When the dice are thrown:

Dale WINS $20 if both show a 6.
Dale WINS $2 if there is only one 6.
He LOSES $2 if neither shows a 6.

Now, because the BOShitheads didn't put the answers up, I can't see if I'm correct. My working was:

(2/12x20)+(1/12x2)+(10/12x-2) = $1.83 financial expectation.

Did I do this right?

what's the question?

Also, i love probability. It's one of the only topics i can get away without studying.

 

[Scanorama]

Probability is all right, the worst in the course would be financial maths, all different formula looks like the same to me

 

[cro_angel]

27(b) best bet is to draw a probability tree then you would see that the probability of not rolling a 6 the first go is 5/6 then on the second go its 5/6 again so you multiply it by each other (so it ends up being 25/36)
(c) FE= 20(1/36) + 2(10/36) + (-2)(25/36)
FE= -$0.277777..
Which is a 28 cent loss.. you need to be multiplying each outcome by its probability of occuring.. then add them. Im not quite sure how you got 2/12, 1/12 and 10/12 :|
and you would be using the probability tree u drew in the first q to answer it..

 

[Sickle]

The next question is:

Dale plays a game with these dice. There is no entry fee. When the dice are thrown:

Dale WINS $20 if both show a 6.
Dale WINS $2 if there is only one 6.
He LOSES $2 if neither shows a 6.

Now, because the BOShitheads didn't put the answers up, I can't see if I'm correct. My working was:

(2/12x20)+(1/12x2)+(10/12x-2) = $1.83 financial expectation.

Did I do this right?

FE= (20 x 1/36)+(2 x 5/36)+(-2 x 25/36)
FE=-$1.16

this is nuts... I hate FE

 

[*bettyboop*]

im with you matt
i hate probablity to
i just dont know wat formula to use or how to use it. it sucks bad.
there wont be heaps of questions on it, so dont freak out. just tryu go better at the better things or things ur good at.
and SICKLE

 

[*bettyboop*]

I Hate Probabilty To

im with you matt
i hate probablity to
i just dont know wat formula to use or how to use it. it sucks bad.
there wont be heaps of questions on it, so dont freak out. just tryu go better at the better things or things ur good at.
and SICKLE is financila expectation part of probablity. ahaha the bos website sucks, its gay how theres no answers just a gay marking thing that sucks anyways... im with you..

 

[Berkay_D]

cro angel is correct

u do (20*1/36) + (2*10/36) - (2*25/36) = .........

 

[JayWalker]

cro angel is correct

u do (20*1/36) + (2*10/36) - (2*25/36) = .........

Answer is here:

Probability he WINS 20 is 1/36 (1/6 * 1/6)
Probability he WINS 2 is 15/36 (1/6 * 5/6)
Probability he LOSES 2 is 25/36 (5/6 * 5/6)

Therefore, you multiply each by their money ammount..

$20 WIN:::::::::$2 WIN::::::::::$2 LOSE (Thats why its -2 not +2)
(20 * [1/36]) + (2 * [15/36]) + (-2 * 25/36)

= $0

Therefore he should expect to walk out as poor as he was when he started


You can see that in my previous post.. Im not sure how you get $1.33 or 28 cents.. according to probability, its Even,

 

[Berkay_D]

the probability of getting (1 SIX) is ....... (5/6*1/6) + (5/6*1/6)

 

[Berkay_D]

which is 10/36.......
i dont get how u get 15/36 for it.

 

[JayWalker]

the probability of getting (1 SIX) is ....... (5/6*1/6) + (5/6*1/6)

Using 2 independant dice,

Probable outcomes are:
Die 1 with the 6 Die 2 with the 6
6:1 1:6
6:2 2:6
6:3 3:6
6:4 4:6
6:5 5:6

Uh, Sec getting back on track (answering this, and 3 other posts AND doing question 10 from HSC paper)...

Yeah its 10/36... My bad!

Recalculating!!

REVISED Answer is here:

Probability he WINS 20 is 1/36 (1/6 * 1/6)
Probability he WINS 2 is 15/36 2(1/6 * 5/6) 2 because it doesnt matter which die the 6 is on
Probability he LOSES 2 is 25/36 (5/6 * 5/6)

Therefore, you multiply each by their money ammount..

$20 WIN:::::::::$2 WIN::::::::::$2 LOSE (Thats why its -2 not +2)
(20 * [1/36]) + (2 * [10/36]) + (-2 * 25/36)

= -10/36

Which means he is likely to walk out 27.77 cents poorer..

Apologies, Cro is correct