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Madison has a choice of two entrees (soup or salad), three main courses (fish, chicken or steak) and three desserts (ice-cream, lemon tart or cheese).

c. Suppose that Madison has the choice to omit the entree and/or the dessert course altogether. Find the probability that:

i. she chooses only two courses.

I have no idea how to work this, I've been trying to 'count' on a tree diagram for hours and I don't know how they get an answer of 15/36.

I know that there are 36 choices because are extra "none" options for the entree and dessert so 3 x 3 x 4.

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The two courses are either entree and main, or main and dessert. The first has six options, the second has nine.

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  • $\begingroup$ Sorry, I don't quite understand it. So there a 6 options for entree and main which are soup, salad, none and fish, chicken and steak. And then the options for main and dessert are fish, chicken, steak and ice cream, lemon tart, cheese and none? Where do you get the 9 from? Could you also use a tree diagram if possible $\endgroup$ – Colin Ritman May 22 at 13:22
  • $\begingroup$ You can't have none because then you would have only one course. $\endgroup$ – Empy2 May 22 at 13:34

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