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    Can I have some help with question 7 on the June 2010 paper? I understand how to do part (i) but not (ii) or (iii).
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    For part (ii), how many possible combinations are there involving boiled rice? Let  \mathbf{A} be the event that they have boiled rice. Then,

     \displaystyle \mathbb{P}\left(\mathbf{A}\right  ) = \frac{\text{ number of combinations involving boiled rice}}{\text{all possible combinations, so part (i)}}

    For part (iii), they can either choose boiled rice or not. You have two situations:

    Situation B: they don't choose boiled rice. How many possible rice combinations do they have now? What about the total number of dish combinations?

    Situation C: they have boiled rice. This means they only have one more rice choice of the remaining three. It also means they cannot choose potatoes, so how many possible vegetable dish combinations do they have now? What about the total number of dish combinations?

    Let me know if you need anymore help
 
 
 
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Updated: February 11, 2017
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