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June 03, 2020, 05:38:32 am

Author Topic: maths  (Read 202 times)

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« on: October 02, 2019, 11:22:58 am »
hi i am doing advanced maths i need help with this question would you be able to break it down for me

A game involves rolling two six-sided dice, followed by rolling a third six-sided
die. To win the game, the number rolled on the third die must lie between the
two numbers rolled previously. For example, if the first two dice show 1 and 4,
the game can only be won by rolling a 2 or 3 with the third die.
(i) What is the probability that a player has no chance of winning before
rolling the third die?
(ii) What is the probability that a player wins the game?


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Re: maths
« Reply #1 on: October 02, 2019, 12:55:30 pm »
Welcome to the forums! :)

There's actually a Mathematics 2U Board you might want to check out a bit later (to post further questions, and to view other questions that you may have that have already been asked! otherwise there are more resources there :) )

For part i), consider this: in what cases do you have no chance of winning before the game starts? Essentially, this maps out a bit like a game of bullshit; if you roll a certain number on the first die, and you roll one up, one down or the same on the second, you instantly have no chance of winning. Drawing a 6x6 table with each outcome helps visualise this best, but you can also list them out case by case :) Try doing this yourself, and then check the spoiler for the solution :)

(1, 1), (1, 2), (2, 1), (2, 2), (2, 3) ... (6, 6) = 16 cases out of a possible 36, ie 4/9

ii) This is a little trickier. There are a few different ways we can consider this;
a) By the number of numbers in between each pair
b) case-by-case (which takes a lot longer)

a) is the easiest option, but you can have a look at b) if you have a lot of time.
You can have anywhere from 1-4 numbers in between each pair ie. for 1, an example (2, 4) and for 4, an example would be (1, 6).
There are eight cases with 1 number in between: hence, we multiply 8/36 (the chance of getting this case) by the chance of hitting a number in between on the third die, which is 1/6.
There are six cases with 2 numbers in between: repeating, we have 6/36 x 2/6
There are four cases with 3 numbers in between: repeating, we have 4/36 x 3/6
There is one case with 4 numbers in between: and lastly we have 2/36 x 4/6

Adding all these probabilities will give you the answer! :)

Also, if I recall correctly, there was a similar question in the 2018 HSC, you might want to check that out! (unless this is from the 2018 HSC?). If you have any further questions, please enquire further! :)

Hope this helps!!
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