Alternative to tree diagrams? (1 Viewer)

toprun91

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Hi,
This is more of a question of curiosity aimed for people who know. But is there an alternative to tree diagrams. How do people calculate the probability of events that require massive tree diagrams that cant be drawn
 

toprun91

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what about tree diagrams the would branch out heaps of times. Ok what if for some unknown reason someone wanted to find the probability of rolling a two 5's and seven 6's when rolling nine dice?. Who can work that out? eh
 

Brontecat

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tbh i can't see that question being asked in a hsc exam, so i wouldn't worry about doing the question too much :)
 

toprun91

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Yer i am not worrying about it i only just wanted to know how people work out those situations in real life
 

Trebla

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Permutations, combinations and use of various probability distributions (e.g. binomial, geometric, multinomial etc)
 

toprun91

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thx everyone. if anyone has spare time and it isnt too hard post how you would solve my hypothetical question above using your 3 unit - 4 unit maths
 

Aquawhite

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In just 2U mathematics (3U uses permutations and factorials) you will only ever get asked short probability questions and often they don't even require a tree diagram (because they very simple and can be done in the head).

At the most the tree will only be 4 times over E.g. something is drawn out three times.... then two options for win or lost on each.
 

toprun91

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In just 2U mathematics (3U uses permutations and factorials) you will only ever get asked short probability questions and often they don't even require a tree diagram (because they very simple and can be done in the head).

At the most the tree will only be 4 times over E.g. something is drawn out three times.... then two options for win or lost on each.
I know, i am just curious to see how its done. I really wanted to do 3 unit but my school is very strict on getting in (you have to be invited) not sure if that's across all schools though. Anyway if anyone wants to show me i would be interested i just want to see if my hypothetical is feasible
 

Aquawhite

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I know, i am just curious to see how its done. I really wanted to do 3 unit but my school is very strict on getting in (you have to be invited) not sure if that's across all schools though. Anyway if anyone wants to show me i would be interested i just want to see if my hypothetical is feasible
My school makes recommendations based your performance... not many people qualify for Extension Maths in my school though... they're under-educated in the field of maths >_> (damn school).

Although, most are encouraged to at least give it a go... see what they think.
 

lychnobity

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what about tree diagrams the would branch out heaps of times. Ok what if for some unknown reason someone wanted to find the probability of rolling a two 5's and seven 6's when rolling nine dice?. Who can work that out? eh
I'm going to be epically wrong, but here goes:

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

toprun91

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My school makes recommendations based your performance... not many people qualify for Extension Maths in my school though... they're under-educated in the field of maths >_> (damn school).

Although, most are encouraged to at least give it a go... see what they think.
I wish i was encouraged i was in a slightly too low maths class to make it into 2 unit so i had to complete all this extra work to get into 2 unit. I also had to get above 65% on the first test or else they would drop me. But i proved them wrong getting 94% and beating all but two 3 unit students. If i had my time over i would love to try harder and do 4 unit. Do you do 3 unit? if so can you show me the way to do the hypothetical probability question
 

lychnobity

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what does the c mean? how did you do it?
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.
 
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danal353

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lol no i wouldn't because i can't bear to even read through it, let alone understand it
 

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