asianese
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- HSC
- 2012
Question/Ponder
Why exactly do we need to 'order' things when doing a binomial probability?
E.g. 1997 HSC MX1
c) In a game of Sic Bo, three regular, six-sided dice are thrown once.
i) In a single game, what is the probability that all three dice show 2?
ii) What is the probability that exactly two of the dice show 2?
iii) What is the probability that exactly two of the dice show the same number?
iv)...
I know now that ii) is a binomial probability, but why do we need the C(3,2)?
If the probability of getting a 2 is (1/6) then getting it twice is (1/36). Then the last outcome needs to be a non-6 so (5/6). But why do we need the C(3,2)? Why do we care about 'ordering' the outcomes? (Answer is C(3,2) * (1/6)^2 * (5/6)
Also for iii), the answer is 6 * (answer in ii). Is the 6 necessary since we could pick any of 1-6 first, so there are 6 outcomes where you can get two of the same?
Many thanks
Why exactly do we need to 'order' things when doing a binomial probability?
E.g. 1997 HSC MX1
c) In a game of Sic Bo, three regular, six-sided dice are thrown once.
i) In a single game, what is the probability that all three dice show 2?
ii) What is the probability that exactly two of the dice show 2?
iii) What is the probability that exactly two of the dice show the same number?
iv)...
I know now that ii) is a binomial probability, but why do we need the C(3,2)?
If the probability of getting a 2 is (1/6) then getting it twice is (1/36). Then the last outcome needs to be a non-6 so (5/6). But why do we need the C(3,2)? Why do we care about 'ordering' the outcomes? (Answer is C(3,2) * (1/6)^2 * (5/6)
Also for iii), the answer is 6 * (answer in ii). Is the 6 necessary since we could pick any of 1-6 first, so there are 6 outcomes where you can get two of the same?
Many thanks