Hello, I have a question on Probability (chapter 9)
Matt 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). Suppose he has the option to omit the entree and/or the dessert altogether. Find the chance that:
a) he chooses all three courses b) he chooses only two courses.
The answer for the first is 1/2 and for the second is 15/36 apparently. I haven't done Probability for such a long time so I'm not really used to it ¯\_(ツ)_/¯
Are you asking how it works that way?
a) Presumably he could choose for entree soup, salad or nothing. The main courses, he could choose three different options but not nothing. He has four options for dessert, ice cream, lemon tart, cheese or nothing.
Thus, he has a 2/3 option of getting entree, a 3/3 option of getting main course, a 3/4 option of getting dessert. Multiply 2/3 and 3/4 together, you get 6/12, which simplifies to 1/2.
b) 1/3 chance of no entree, 1/4 chance of no dessert. Don't multiply them together here or that would be finding when he would have only one course. So I think what you want there is multiplying no entree and dessert. Then multiply entree and no dessert. Then add the probabilities together. There may be an easier way of doing this.
no entree and dessert: 1/3 * 3/4 = 3/12 = 1/4
entree and no dessert: 2/3 * 1/4 = 2/12 = 1/6
1/4 + 1/6 = 6/24 + 4/24
= 10/24 = 5/12
Because there are three options of main course, we have 3/3 gets this successfully.
Multiply 3/3 by 5/12
= 15/36
Hope this was what you wanted... hope it's right, too, I haven't done Probability for ages. Maybe someone else should check.