dont know it
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#1
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#1
https://prnt.sc/jdv22b

^^ screenshot of question
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the bear
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#2
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so you have a normal dice:

https://cdn.shopify.com/s/files/1/17...g?v=1501628890

what is the chance it lands on a 6 ?
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dont know it
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(Original post by the bear)
so you have a normal dice:

https://cdn.shopify.com/s/files/1/17...g?v=1501628890

what is the chance it lands on a 6 ?
1/6. The answer says something specifically about it being a cubical dice and I have no idea what difference it makes. The answer says a cubical die has 18 sides.. so yeah I'm really confused.
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the bear
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(Original post by dont know it)
1/6. The answer says something specifically about it being a cubical dice and I have no idea what difference it makes. The answer says a cubical die has 18 sides.. so yeah I'm really confused.
a cubical die is just a normal 6 sided cube.
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ghostwalker
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(Original post by dont know it)
The answer says a cubical die has 18 sides..
Not in this universe. The number of sides is 6. 6 is the number of sides. The number of sides is not 5 - 5 is too small - nor is the number of sides 7, for 7 is way too big [/Pythonesque]

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dont know it
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(Original post by the bear)
a cubical die is just a normal 6 sided cube.
Agreed. This is what the answer says though: https://prnt.sc/jdv896
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the bear
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they are giving an example of an INCORRECT answer.

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dont know it
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(Original post by the bear)
they are giving an example of an INCORRECT answer.

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Oh I see. I still don't understand why it's not 1/6 though .
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the bear
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(Original post by dont know it)
Oh I see. I still don't understand why it's not 1/6 though .
it IS 1/6 if you do it correctly. they are supposing that some doofus thinks a dice has 18 sides.
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(Original post by the bear)
it IS 1/6 if you do it correctly. they are supposing that some doofus thinks a dice has 18 sides.
Oh haha. Never seen a mark scheme do that, my bad. Thanks for the help.
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Johnw77
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It can be simplified greatly by considering what events must occur in each of the 4 cases.

For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.

Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)

In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.
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Johnw77
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(Original post by Johnw77)
It can be simplified greatly by considering what events must occur in each of the 4 cases.

For example, for Chris to win on his first roll, we need the first two people to NOT win, and for the third person TO win. The probability of any roll resulting in a six is 1/6, so the probability of a win is 1/6 and the probability of a loss is 5/6. so the probability that the first roll is a loss is 5/6, the second roll is a loss is also 5/6, and the third roll is a win is 1/6. Multiply these to get 25/216, which is your answer for the first part.

Similarly for the next three parts, just think how many losses you need before you need a win. e.g. say you want to find the probability of winning on the 5th roll. This has probability (5/6)^4 x (1/6) = 625/7976, because we need exactly 4 losses and exactly 1 win, in that order. (order is important, we win MUST be the last roll, as the game ends then)

In case you didn't know, as I can't remember if this is on the S1 course, (if it is and you are familiar with it I apologise if this sounds patronising) this is an example of a geometric distribution, which models the probability of it taking a certain number of attempts to achieve a success once, and then stopping. It is called geometric because every time we increase the number of trials before a success, the probability decreases by the same factor, just like geometric sequences.
I should say the above technique on applies easily to the first 2 parts. The last two parts are slightly trickier, because each has multiple possibilities to still be true. For example, for Bill to throw the die exactly three times, we need to win on or after his third throw, but before his fourth throw, so there are actually 4 possible ways in which this can occur. You just need to find them, and work out the probability of each, and add them up.
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