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1. Start by filling x=1. That's simply the chance of rolling a one, then to the power of 4 because there are four rolls (I assume you're familiar with this).

You can then do x=4 by taking the chance of rolling any number less than or equal to 4, then raising it to the power of 4. However, you will have to take into account the values for x=1, 2 and 3 which would have been double-counted, so you'll have to subtract those.

Of course when you're down to just one box left, you can find it by just doing 1 minus everything else.
2. (Original post by TheMindGarage)
Start by filling x=1. That's simply the chance of rolling a one, then to the power of 4 because there are four rolls (I assume you're familiar with this).

You can then do x=4 by taking the chance of rolling any number less than or equal to 4, then raising it to the power of 4. However, you will have to take into account the values for x=1, 2 and 3 which would have been double-counted, so you'll have to subtract those.

Of course when you're down to just one box left, you can find it by just doing 1 minus everything else.
Thank you, you wouldn't also happen to know how to work out 18 aswell would you?
3. (Original post by Alexia_17)
Thank you, you wouldn't also happen to know how to work out 18 aswell would you?
Are you familiar with the binomial distribution? Essentially it's:

(probability of rolling a 6)3 x (probability of not rolling a 6)7-3 x (number of possible ways to roll 3 sixes in 7 rolls)

For that last part of the formula, you need to use the combinatoric function (the capital C with a subscript number and a superscript number).
4. (Original post by TheMindGarage)
Are you familiar with the binomial distribution? Essentially it's:

(probability of rolling a 6)3 x (probability of not rolling a 6)7-3 x (number of possible ways to roll 3 sixes in 7 rolls)

For that last part of the formula, you need to use the combinatoric function (the capital C with a subscript number and a superscript number).
Thank you so much. Just checking, would the last part be 7C3?
5. (Original post by Alexia_17)
Thank you so much. Just checking, would the last part be 7C3?
Yes.

Part B should be very straightforward once you have an answer for A.
6. (Original post by TheMindGarage)
Yes.

Part B should be very straightforward once you have an answer for A.
Would you just multiply the answer from part A by 1/6 (the probability of getting a six)?
7. (Original post by Alexia_17)
Would you just multiply the answer from part A by 1/6 (the probability of getting a six)?
Sorry for late reply, but yes.

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