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# Maths A level Questions watch

1. I will post some questions. I already tried attempting them, so please just give me the solutions. THANK YOU
2. Post away. *Prays it's not FP3*
I will post some questions. I already tried attempting them, so please just give me the solutions. THANK YOU
4. (Original post by Bulletzone)
Post away. *Prays it's not FP3*
it is s1 and m2
it is s1 and m2
Hate stats. Will attempt m2.

Which part, ii)?
Some working would be nice. and no
the following answers according to the textbook is:

i.2/27
ii.125/216
iii.5/9
Some working would be nice. and no
the following answers according to the textbook is:

i.2/27
ii.125/216
iii.5/9
All dice are fair right?
10. (Original post by thekidwhogames)
Which part, ii)?
part i and iii
11. This is A-level? Seems like GCSE. Is this AS?
12. (Original post by Bulletzone)
Hate stats. Will attempt m2.
Good.
13. Part ii:

P(6') = 1-P(6) = 5/6.
Hence 5/6 cubed = 125/216 (as three rolls).
Some working would be nice. and no
the following answers according to the textbook is:

i.2/27
ii.125/216
iii.5/9
Ahh i realised my error. Part ii is simply 5/6 cubed, which is the probabilty of not obtaining a six when a dice is thrown. Would you like workings fr the other 2 parts?
For part (i), think about the probability of getting all three sixes, and then the probabilities of each of the three ways of getting exactly two sixes; add the results together.

For part (ii), find the probability of not getting a six with one of the dice, and then cube it.

For part (iii), I would try to think of all combinations of how you can get at least two numbers the same, and their probabilities together, and then subtract this value from 1.

I hope that this helps!

EDIT: Actually, my method for part (iii) is overly long-winded. beckyyfoster did a much better solution three posts down.

Spoiler:
Show

(6/6) * (5/6) * (4/6) = 1 * (20/36) = 5/9
Ahh i realised my error. Part ii is simply 5/6 cubed, which is the probabilty of not obtaining a six when a dice is thrown. Would you like workings fr the other 2 parts?

Show me all the working for all the parts please. thank you
17. (Original post by thekidwhogames)
Part ii:

P(6' = 1-P(6) = 5/6.
Hence 5/6 cubed = 125/216 (as three rolls).
thx
18. okay so:

ii) probability of no 6's means throwing a 1/2/3/4/5 on all 3 throws
therefore: 5/6 * 5/6 * 5/6 = (5/6)^3 = 125/216

iii) different scores on all dice mean throwing any on 1st throw, so then there are only 5 possibilities for 2nd and 4 for third
therefore: 1 * 5/6 * 4/6 = 20/36 = 5/9

Show me all the working for all the parts please. thank you
Part i) it is simply the probability of getting two sixes, which can happen 3 different ways plus the probability of getting 3 sixes because it says at least two sixes meaning 2 or greater. So you get 3x (1/6 x1/6 x5/6) + (1/6)^3.
Part iii is simply 1/6 x 5/6 ( because you have 5 remaining digits to choose from) x4/6 ( because you have 4 remaining digits to choose from). This can happen 6 different ways so we multiply by 6.
20. hh
(Original post by beckyyfoster)
okay so:

ii) probability of no 6's means throwing a 1/2/3/4/5 on all 3 throws
therefore: 5/6 * 5/6 * 5/6 = (5/6)^3 = 125/216

iii) different scores on all dice mean throwing any on 1st throw, so then there are only 5 possibilities for 2nd and 4 for third
therefore: 1 * 5/6 * 4/6 = 20/36 = 5/9
Thank you very much for the third part

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