Alternative to tree diagrams? (1 Viewer)

[blaze.bluetane]

Are you sure? I dunno...

Does it work just multiplying the two probabilities?

For EXT1ers, I think it works out like this;

"If there is a series of repeated independent events, where the probability of success is p and the probability of failure is q, the probability of successes in n trials is given by the terms in the expansion of (p+q)ⁿ"

EXT1 students are able to expand any binomial expression, recalling that the probability of r successes in n events is given by;
P(E)=ⁿC_r.p^r.q^n-r

For calculating the probability of two '5's;
let P(two '5's) =P(A)
= ⁹C₂.(1/6)².(5/6)⁷
= 0.279... or (2812500/10077696)

For calculating the probability of seven '6's;
let P(seven '6's)=P(B)
= ⁹C₇.(1/6)⁷.(5/6)²
= 0.0000893... or (25/279936)

Therefore, when A and B both occur, P(E)=P(AB)

P(AB)=P(A).P(B)
=0.0000249...

For 2Units, ⁿC_r= [ n(n-1)(n-2)(n-3)...(n-r+1) ] / [ 1 x 2 x 3 x 4 x 5 x ... x r ]
⁹C₂= [9 x 8 ]/[1 x 2]
= 36 ⁹C₇= [9 x 8 x 7 x 6 x 5 x 4 x 3]/[1 x 2 x 3 x 4 x 5 x 6 x 7]
= 36


lychnobity it looks like you have not multiplied by the probability of failure part
I am very new to this so help me out.

P.s. in any case toprun91, your fives and sixes are not happening any time soon

 

[Brontecat]

The C, (if you look on your calculator should be on top of some button as "nCr") is a function that calculates the number of ways of selecting r objects from a total of n objects.

Technically, what my working out should look like is:

⁹C₂ x (1/6)<SUP>2</SUP> x ⁹C₇ x (1/6)<SUP>7</SUP> = 1/7776

Take ⁹C₂ for example, the 9 represents the total no. of objects (dice), but I wanted to choose 2, so ⁹C₂ calculates the number of ways I can choose 2 objects out of 9 objects.

The ⁹C₂ chooses any 2 dice to have a result of 5, and the (1/6)<SUP>2</SUP> is the probability of rolling 2 fives.

The same applies for the ⁹C₇, any 7 die to have a 6, and (1/6)<SUP>7 </SUP>is the probability of rolling 7 sixes.

i had no idea that was 3U
my teacher taught me that back in term 1 this year- i thought it was a bit hard
thanks for the explanation

 

[blaze.bluetane]

I really am unsure as to whether I am on the right track or not.
Someone care to shed some light?

Is that how you properly calculate the probabilities and to get the "two '5's and seven '6's" result do you multiply the probabilities?

 

[toprun91]

Are you sure? I dunno...

Does it work just multiplying the two probabilities?

For EXT1ers, I think it works out like this;

"If there is a series of repeated independent events, where the probability of success is p and the probability of failure is q, the probability of successes in n trials is given by the terms in the expansion of (p+q)ⁿ"

EXT1 students are able to expand any binomial expression, recalling that the probability of r successes in n events is given by;
P(E)=ⁿC_r.p^r.q^n-r

For calculating the probability of two '5's;
let P(two '5's) =P(A)
= ⁹C₂.(1/6)².(5/6)⁷
= 0.279... or (2812500/10077696)

For calculating the probability of seven '6's;
let P(seven '6's)=P(B)
= ⁹C₇.(1/6)⁷.(5/6)²
= 0.0000893... or (25/279936)

Therefore, when A and B both occur, P(E)=P(AB)

P(AB)=P(A).P(B)
=0.0000249...

For 2Units, ⁿC_r= [ n(n-1)(n-2)(n-3)...(n-r+1) ] / [ 1 x 2 x 3 x 4 x 5 x ... x r ]
⁹C₂= [9 x 8 ]/[1 x 2]
= 36 ⁹C₇= [9 x 8 x 7 x 6 x 5 x 4 x 3]/[1 x 2 x 3 x 4 x 5 x 6 x 7]
= 36


lychnobity it looks like you have not multiplied by the probability of failure part
I am very new to this so help me out.

P.s. in any case toprun91, your fives and sixes are not happening any time soon

Could you explain it in a way a 2 unit person would understand

 

[lychnobity]

Could you explain it in a way a 2 unit person would understand

blaze.bluetane is correct, originally my working out should look like:

⁹C₂ x (1/6)² x (5/6)⁷ x ⁹C₇ x (1/6)⁷x (5/6)²

The explanation for the ⁹C₂ x (1/6)² part is exactly the same, but you have to multiply the probability of tossing two 5s by the probability of NOT getting two 5s. Call the probability of getting a 5 F, and the probability of not getting a 5, N.

When you roll the die 9 times, and want the probability of getting two 5s, you will get results that look like this:

FFN,NNN,NNN

From the above, I'd have to account for the probability of the 7 die NOT getting 5s, so that's why you have to multiply it by (5/6)⁷

Similarly, the explanation for ⁹C₇ x (1/6)⁷ is the same. But you would have to multiply that result by (5/6)² to account for the probability of 2 of the die to NOT get a 6.

 

[toprun91]

blaze.bluetane is correct, originally my working out should look like:

⁹C₂ x (1/6)² x (5/6)⁷ x ⁹C₇ x (1/6)⁷x (5/6)²

The explanation for the ⁹C₂ x (1/6)² part is exactly the same, but you have to multiply the probability of tossing two 5s by the probability of NOT getting two 5s. Call the probability of getting a 5 F, and the probability of not getting a 5, N.

When you roll the die 9 times, and want the probability of getting two 5s, you will get results that look like this:

FFN,NNN,NNN

From the above, I'd have to account for the probability of the 7 die NOT getting 5s, so that's why you have to multiply it by (5/6)⁷

Similarly, the explanation for ⁹C₇ x (1/6)⁷ is the same. But you would have to multiply that result by (5/6)² to account for the probability of 2 of the die to NOT get a 6.

What does the ⁹C₂mean? Is there some kind of formula i can put these kind of questions into 

 

[lychnobity]

What does the ⁹C₂mean? Is there some kind of formula i can put these kind of questions into 

Take ⁹C₂ for example, the 9 represents the total no. of objects (dice), but I wanted to choose 2, so ⁹C₂ calculates the number of ways I can choose 2 objects out of 9 objects.

The general form for this "C" formula is ⁿC_r

where n is the total number of objects
---and r is the number of objects you would like to choose.

For example in your hypothetical question, you wanted to choose out of 9 die, 2 to always have a result of five.

To find the number of ways that this can happen, you put the numbers into the formula ie ⁹C<sub>2

</sub>Say you had 4 apples, and you wanted to choose 2 of them.

Name the apples A,B,C,D

So in order to choose 2 of them, you can have these combinations:

AB
AC
AD
BC
BD
CD

ie a total of 6.

BUT, if you didn't want to come up with all of the combinations, you could have just used the formula: ⁿC_r. In this case your formula would be ⁴C₂ = 6

 

[Brontecat]

do u think it would be worth it for a 2U student to learn Combinations (ⁿC_r)and the binomial probability distribution (P(r successes) = ⁿC_{r[/sup]p^rq^n-r) or would it just be a waste of time?}

 

[oly1991]

do u think it would be worth it for a 2U student to learn Combinations (ⁿC_r)and the binomial probability distribution (P(r successes) = ⁿC_{r[/sup]p^rq^n-r) or would it just be a waste of time?}

<sub>it wouldn't hurt.

sometimes you can make some silly errors in tree diagrams and you can use the more complex formulas to back up your answers....thats what i do</sub>

 

[blaze.bluetane]

Big waste of time. No doubt about it....

However, if you intend to study further probability later in life, I suppose you could give it a go if you really wanted. But remember, time spent learning that stuff is time that could have been spent learning stuff that is in your HSC!

 

[blaze.bluetane]

Could you explain it in a way a 2 unit person would understand

For 2 Unit students;

P(two '5's)

To find the probability of obtaining two '5's, we multiply the probabilities along the tree [or a branch of the tree] to obtain every possible way to obtain the result.

So, if I want two '5's, will have to achieve these fives somewhere among the nine dice rolls. So for example, I could roll a five on the first and second rolls; I will notate this by (1,2), noting that this is identical to rolling a five on the second and first rolls, (2,1);

So the sample space is;
(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(1,9)
(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)
(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)
(4,5)(4,6)(4,7)(4,8)(4,9)
(5,6)(5,7)(5,8)(5,9)
(6,7)(6,8)(6,9)
(7,8)(7,9)
(8,9)

Which is a total of 36 outcomes (This is where ⁹C₂ and ⁹C₇ come from)

And then we multiply this by the probability of obtaining this outcome, which is the same, for example, for rolling [5,5,3,1,2,3,6,1,3] or [5,3,3,6,5,1,3,1,2] and is;

(1/6)(1/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)=(1/6)^2 x (5/6)^7

This is any branch of the probability tree where '5' occurs exactly twice in nine rolls

which is the same probability, wherever '5' occurs

So we have 36 different possible ways of getting two '5's, each with a probability of occuring of (1/6)^2 x (5/6)^7

--------------------------------------------------------------------------------------------------------------------------------------

P(seven '6's)

If we consider getting seven '6's, there are also 36 different ways that this can occur, where the previous table can be reinterpreted as the dice where there is no '6' instead of obtaining a five.

In this light, the probability of getting seven '6's, is;

(1/6)(1/6)(1/6)(1/6)(1/6)(1/6)(1/6)(5/6)(5/6)=(1/6)^7 x (5/6)^2

So there are 36 times when we can get seven '6's and they are each equally probable; with a probability of (1/6)^7 x (5/6)^2

--------------------------------------------------------------------------------------------------------------------------------------

P(two '5's and seven '6's)

When two events both occur P(a) and P(b), then then the probability is calculated by multiplying the probabilities;


P(E)= 36 x (1/6)^7 x (5/6)^2 x 36 x (1/6)^2 x (5/6)^7

 

[toprun91]

For 2 Unit students;

P(two '5's)

To find the probability of obtaining two '5's, we multiply the probabilities along the tree [or a branch of the tree] to obtain every possible way to obtain the result.

So, if I want two '5's, will have to achieve these fives somewhere among the nine dice rolls. So for example, I could roll a five on the first and second rolls; I will notate this by (1,2), noting that this is identical to rolling a five on the second and first rolls, (2,1);

So the sample space is;
(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(1,9)
(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)
(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)
(4,5)(4,6)(4,7)(4,8)(4,9)
(5,6)(5,7)(5,8)(5,9)
(6,7)(6,8)(6,9)
(7,8)(7,9)
(8,9)

Which is a total of 36 outcomes (This is where ⁹C₂ and ⁹C₇ come from)

And then we multiply this by the probability of obtaining this outcome, which is the same, for example, for rolling [5,5,3,1,2,3,6,1,3] or [5,3,3,6,5,1,3,1,2] and is;

(1/6)(1/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)=(1/6)^2 x (5/6)^7

This any the branch of the probability tree where '5' occurs exactly twice in nine rolls

which is the same probability, wherever '5' occurs

So we have 36 different possible ways of getting two '5's, each with a probability of occuring of (1/6)^2 x (5/6)^7

--------------------------------------------------------------------------------------------------------------------------------------

P(seven '6's)

If we consider getting seven '6's, there are also 36 different ways that this can occur, where the previous table can be reinterpreted as the dice where there is no '6' instead of obtaining a five.

In this light, the probability of getting seven '6's, is;

(1/6)(1/6)(1/6)(1/6)(1/6)(1/6)(1/6)(5/6)(5/6)=(1/6)^7 x (5/6)^2

So there are 36 times when we can get seven '6's and they are each equally probable; with a probability of (1/6)^7 x (5/6)^2

--------------------------------------------------------------------------------------------------------------------------------------

P(two '5's and seven '6's)

When two events both occur P(a) and P(b), then then the probability is calculated by multiplying the probabilities;


P(E)= 36 x (1/6)^7 x (5/6)^2 x 36 x (1/6)^2 x (5/6)^7

WOW! thanks heaps for that i cant believe i actually understood that you should get into the study guide writing business!. I just had one more question in my hypothetical i meant as in the probability of a rolling stuff like [5,5,6,6,6,6,6,6,6] , or [5,6,6,6,6,6,6,6,5] meaning you couldn't roll other numbers. Is what you showed me how you would work out that?. Also thx heaps as well to lynchnobity

 

[blaze.bluetane]

What I have done in the 2 unit is work out three probabilities...

First, P(two '5's), then P(seven '6's)....

When doing probability, if you want to find the probability of two particular things happening P(a) and P(b) both together, then you must multiply their particular probabilities... notated {P(a) and P(b) = P(a.b)}

So we get these two individual probabilities and multiply.

P(two '5's) and P(seven '6's) = P(two '5's) x P(seven '6's)

i.e. P(3,4,5,1,3,2,1,5,6) x P(6,6,4,6,6,6,6,1,6) = P(6,6,5,6,6,6,6,5,6)

So your question was a combo of two conditions really... the '5's and the '6's, and each probability had to be calculated separately then combined

The EXT1 way is really shorthand for all this so that it sounds more complicated than it is. EXT1 students should recognise this, however it is usually a 'learn the formula and apply it' sort of thing.

I am relieved that you understood the previous post... I thought it was a bit dodgy. Oh well

hope this helps...

 

[toprun91]

What I have done in the 2 unit is work out three probabilities...

First, P(two '5's), then P(seven '6's)....

When doing probability, if you want to find the probability of two particular things happening P(a) and P(b) both together, then you must multiply their particular probabilities... notated {P(a) and P(b) = P(a.b)}

So we get these two individual probabilities and multiply.

P(two '5's) and P(seven '6's) = P(two '5's) x P(seven '6's)

i.e. P(3,4,5,1,3,2,1,5,6) x P(6,6,4,6,6,6,6,1,6) = P(6,6,5,6,6,6,6,5,6)

So your question was a combo of two conditions really... the '5's and the '6's, and each probability had to be calculated separately then combined

The EXT1 way is really shorthand for all this so that it sounds more complicated than it is. EXT1 students should recognise this, however it is usually a 'learn the formula and apply it' sort of thing.

I am relieved that you understood the previous post... I thought it was a bit dodgy. Oh well

hope this helps...

Thanks heaps that last bit of info will help me a lot . I know you have done alot to help me but if you have time how would you do really large replacement in equation from. Say in a hypothetical you 20 blue marbles and and 30 red marbles and you take out 9 marbles what is the probability of getting 2 blue marbles and 7 red marbles.

 

[toprun91]

Does anyone know?

 

[Timothy.Siu]

Thanks heaps that last bit of info will help me a lot . I know you have done alot to help me but if you have time how would you do really large replacement in equation from. Say in a hypothetical you 20 blue marbles and and 30 red marbles and you take out 9 marbles what is the probability of getting 2 blue marbles and 7 red marbles.

i have no idea lol.
we could do it if the probability of getting a blue marble was 2/5 and the probability of a red one was 3/5.

 

[annabackwards]

i have no idea lol.
we could do it if the probability of getting a blue marble was 2/5 and the probability of a red one was 3/5.

That is what i'd do.

 

[blaze.bluetane]

Thanks heaps that last bit of info will help me a lot . I know you have done alot to help me but if you have time how would you do really large replacement in equation form. Say in a hypothetical you have 20 blue marbles and 30 red marbles and you take out 9 marbles what is the probability of getting 2 blue marbles and 7 red marbles.

I have a bit of a love-hate relationship with probability...

It is never a learn the formula and substitute in the values sort of thing. You always have to think about it...

ARE THE MARBLES REPLACED? "really large replacement in equation form"

If so,

P(2 Blue)= (2/5)² x (3/5)⁷
and there are ⁹C₂ possible ways of achieving this outcome
P(7 Red)= (3/5)⁷ x (2/5)²
and there are ⁹C₇ possible ways of achieving this outcome

Notice that..
1. ⁹C₇=⁹C₂
That is, ⁿC_r=ⁿC_n-r
This result comes from the symmetry of Pascal's triangle FYI

2. This is the same situation with a different probability

Oh of course...
P(2 blue and 7 red)= [⁹C₂]² x (2/5)⁹ x (3/5)⁹
Whatever that turns out to be

This is based on the situation where 9 marbles are drawn out, one after the other, and replaced after their colour is observed.

hope this helps
bb

p.s. sorry for the late reply

 

[Aquawhite]

There's just a substitute into formula kinda thing in probability/perms/combs? I always have to think about them.

 

[toprun91]

I have a bit of a love-hate relationship with probability...

It is never a learn the formula and substitute in the values sort of thing. You always have to think about it...

ARE THE MARBLES REPLACED? "really large replacement in equation form"

If so,

P(2 Blue)= (2/5)² x (3/5)⁷
and there are ⁹C₂ possible ways of achieving this outcome
P(7 Red)= (3/5)⁷ x (2/5)²
and there are ⁹C₇ possible ways of achieving this outcome

Notice that..
1. ⁹C₇=⁹C₂
That is, ⁿC_r=ⁿC_n-r
This result comes from the symmetry of Pascal's triangle FYI

2. This is the same situation with a different probability

Oh of course...
P(2 blue and 7 red)= [⁹C₂]² x (2/5)⁹ x (3/5)⁹
Whatever that turns out to be

This is based on the situation where 9 marbles are drawn out, one after the other, and replaced after their colour is observed.

hope this helps
bb

p.s. sorry for the late reply

Thank you i really appreciate you responses. I am sorry but i think i worded my question wrong. What i meant to day was without replacement. If you could help me solve this i will really appreciate it thx