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Genrish500 [490]
3 years ago
7

Suppose that a group of 10 students are to be assigned to four dorm rooms: a single, a double, a triple and a quad. How many way

s are there are making the assignments? Choose the best answer.
Mathematics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

Total number of ways to select all sstudent to make assignment = 10×36×70×1 = 25200

Step-by-step explanation:

We have given number of students = 10

These 10 students are assigned to four dorm rooms a single, a double, a triple and a quad

So number of ways to assigned single student =^{10}C_1=\frac{10!}{1!9!}=10\\

Now left student = 10 -1 = 9

So number of ways to assigned double student =^{9}C_2=\frac{9!}{2!7!}=36

Now left student = 9-2 = 7

So number of ways to assigned triple student =^{7}C_3=\frac{7!}{2!4!}==70

Now left student = 7 - 3 = 4

So number of ways to assigned quad student =^{4}C_4=\frac{4!}{4!0!}=1

So total number of ways to select all sstudent to make assignment = 10×36×70×1 = 25200

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baherus [9]

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

p = 0.553, n = 10

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

In which

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

P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003

P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039

P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219

P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724

P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

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Answer:

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