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Yakvenalex [24]
3 years ago
8

To fulfill the requirements for a certain degree, a student can choose to take any 7 out of a list of 20 courses, with the const

raint that at least 1 of the 7 courses must be astatistics 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
Mathematics
1 answer:
s2008m [1.1K]3 years ago
4 0

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 \binom{5}{1}\binom{19}{6}

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: \binom{20}{7}, but now we should subtract from that all the possibilities when none of the courses chose are a statistic course, that's is \binom{15}{7} 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:

\binom{20}{7}-\binom{15}{7}=71085

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 \binom{5}{1}\binom{19}{6} means) because the selection of one course affect the other choices.

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Answer:

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Step-by-step explanation:

Let <em>X</em> = number of consumer's who recognize the brand.

The probability of the random variable <em>X</em> is, P (X) = <em>p</em> = 0.40.

A random sample of size, <em>n</em> = 6 consumers are selected.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

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(a)

Compute the value of P (X = 5) as follows:

P(X=5)={6\choose 5}0.40^{5}(1-0.40)^{6-5}\\=6\times0.01024\times0.60\\=0.036864\\\approx0.0369

Thus, the probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.

(b)

Compute the value of P (X = 6) as follows:

P(X=6)={6\choose 6}0.40^{6}(1-0.40)^{6-6}\\=1\times 0.004096\times1\\=0.004096\\\approx0.0041

Thus, the probability that all of the selected consumers recognize the brand name is 0.0041.

(c)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = P (X = 5) + P (X = 6)

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Thus, the probability that at least 5 of the selected consumers recognize the brand name is 0.041.

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An event is considered unusual if the probability of its occurrence is less than 0.05.

The probability of 5 customers recognizing the brand name is 0.0369.

This probability value is less than 0.05.

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