Mathematics, 18.06.2020 15:57, mdakane3772

A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calculations and reasoning for the following: a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course? b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course? c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that All 4 students selected took online courses

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answered: charlesiarenee0

a) 60% probability that student took at least one online course

b) 40% probability that student did not take an online course

c) 12.96% probability that all 4 students selected took online courses.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

3 out of 5 students enrolled in higher education took at least one online course last fall.

This means that $p = \frac{3}{5} = 0.6$

a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?

This is P(X = 1) when n = 1. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{1,1}.(0.6)^{1}.(0.4)^{0} = 0.6$

60% probability that student took at least one online course.

b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?

This is P(X = 0) when n = 1.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1,1}.(0.6)^{0}.(0.4)^{1} = 0.4$

40% probability that student did not take an online course

c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses

This is P(X = 4) when n = 4.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296$

12.96% probability that all 4 students selected took online courses.

answered: Guest

The answer would be b ! explanation: the formula for sa *surface area * is 2×l×w+2×l×h+2×w×h so in this case l=4w=12h=8so you use the formula to solve it and it give you the answer of 352^cm^2
Someone me. it would be nice if u could also explain how u ot the answer and how u do the equation

Someone me. it would be nice if u could also explain how u ot the answer and how u do the equation

answered: Guest

the answer is mp

step-by-step explanation:

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