please check my answers and problem at the end , too!
the cafeteria in the school offers 3 side items with the meal to the kids, among them: A. fries, B. carrots, C. macaroni, D broccoli, E. beans.
how many choices each child have?
- i think it is 10 because the order does not matter, as long as the 3 items are the same (like ABC and CAB count as the same..
what is the probability that a child will choose macaroni, carrots and beans? (BCE)
- all the possibilities are 10, this combination is 1 out of 10
and the problem:
what is a probability that a child will choose fries?
- normally i would say there is 5 side items, so 1 out of 5 is the chance, but if we check the total choices (the 3-groups) out of the 10 choice variations there is 6 has the fries in it.. so which is it, 6/10 or 1/5????
To answer your first question, yes, you are correct that the order of the side items doesn't matter. So, if a child needs to choose 3 side items out of the 5 options, there are 10 different combinations they can choose (e.g., ABC, ACB, BAC, BCA, etc.). This is because you can think of it as choosing one item at a time without replacement (meaning once an item is chosen, it cannot be chosen again), so the calculation is similar to finding combinations without repetition.
For the second question, you correctly stated that there are 10 different possibilities for the child to choose their side items. Therefore, the probability that they choose macaroni, carrots, and beans (BCE) is indeed 1 out of 10.
Now let's address the problem regarding the probability of a child choosing fries. If we consider the total number of different choices (10), you're correct in saying that there are 6 combinations out of the 10 choices that include fries. However, if we consider the total number of side items (5), then the probability of choosing fries would be 1 out of 5.
Both approaches are valid, but they involve different ways of defining the probability. If we define the probability based on the total number of different choices (10), the probability of choosing fries would be 6 out of 10, or 6/10. On the other hand, if we define the probability based on the total number of side items (5), then the probability would be 1 out of 5, or 1/5.
The choice of which approach to use depends on how you define the sample space. If you consider the set of all possible choices (10 choices), then it would be 6/10. If you consider the set of all possible side items (5 side items), then it would be 1/5.