Complete answer:
Fulfill the requirements for a certain degree, a student can choose to take any 7 out of a list of 20 courses, with the constraint that at least 1 of the 7 courses must be a statistics course. Suppose that 5 of the 20 courses are statistics courses.
(a) How many choices are there for which 7 courses to take?
(b) Explain intuitively why the answer to (a) is not [tex] \binom{5}{1}\binom{19}{6} [/tex]
Answer:
a) 71085 choices
b) See below
Step-by-step explanation:
a) First we're going to calculate in how many ways you can take 7 courses from a list of 20 without the constraint that at least 1 of the 7 courses must be a statistics course, that's simply a combination of elements without repetition so it's: [tex]\binom{20}{7} [/tex], but now we should subtract from that all the possibilities when none of the courses chose are a statistic course, that's is [tex]\binom{15}{7} [/tex] because 15 courses are not statistics and 7 are the ways to arrange them. So finally, the choices for which 7 courses to take with the constraint that at least 1 of the 7 courses must be a statistics course are:
[tex]\binom{20}{7}-\binom{15}{7}=71085 [/tex]
b) It's important to note that the constraint at least 1 of the 7 courses must be a statistics course make the possible events dependent, we can not only fix an statistic course and choose the others willingly ( that is what [tex] \binom{5}{1}\binom{19}{6} [/tex] means) because the selection of one course affect the other choices.
In this exercise, we have to use our knowledge of statistics to calculate how many options can be chosen for a course, so we find that:
a) 169 choices
b) 1 of the 7 courses must be a statistics course, because the selection of one course affect the other choices.
So from the data reported in the exercise we can say that:
a) First we're going to calculate in how many ways you can take 7 courses from a list of 20 without the constraint that at least 1 of the 7 courses must be a statistics course, that's simply a combination of elements without repetition so it's. So we have that;
[tex]C=\frac{m!}{p!(m-p)!}\\C=\frac{20!}{7!(20-7)!}\\C=169[/tex]
b) It's influential to note that the restraint not completely 1 of the 7 courses must be a enumeration course form the attainable occurrence determined by, we can not only fix a statistic course and select the possible choice gladly because the preference from among choices of individual course influence the added selection.
See more about statistics at brainly.com/question/10951564
