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Degger [83]
3 years ago
8

There are 17 portable mini suites (a.k.a. cages) in a row at the paws and claws holiday pet resort. they are neatly labeled with

their guests' names. there are 8 poodles and 9 tabbies. how many ways can the "suites" be arranged if:
Mathematics
1 answer:
DiKsa [7]3 years ago
4 0
Given that there are 17 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with their guests' names. There are 8 poodles and 9 tabbies. How many ways can the "suites" be arranged if: a) there are no restrictions.

b) cats and dogs must alternate.

c) dogs must be next to each other.

d) dogs must be next to each other and cats must be next to each other.


Part A:

If there are no restrictions, then the number of ways the "suites" can be arranged is given by:

17!=17\times16\times15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1=355,687,428,096,000 \ ways



Part B:

If cats and dogs must alternate, then the number of ways the "suites" can be arranged is given by:

9!\times8!=9\times8\times7\times6\times5\times4\times3\times2\times1\times8\times7\times6\times5\times4\times3\times2\times1=14,631,321,600 \ ways



Part C:

If dogs must be next to each other, then the group of dogs are taken as 1 object and there are now 10 objects to arrange. The number of ways to arrange 10 objects is given by:

10!=10\times9\times8\times7\times6\times5\times4\times3\times2\times1=3,628,800
 \ ways

Also, the group of dogs can be arranged in:

8!=8\times7\times6\times5\times4\times3\times2\times1=40,320
 \ ways

Therefore, the total number of ways the "suites" can be arranged is given by:

10!\times8!=3,628,800\times40,320=146,313,216,000 \ ways



Part D:

If dogs must be next to each other and cats must be next to each other, then the group of dogs are taken as 1 object and the group of cats are taken as 1 object, then there are now 2 objects to arrange.

The number of ways to arrange two objects is given by:

2!=2\times1=2

The number of ways the 9 dogs are to be arranged is given by:

9!=9\times8\times7\times6\times5\times4\times3\times2\times1=362,880

The number of ways the 8 cats are to be arranged is given by:
8!=8\times7\times6\times5\times4\times3\times2\times1=40,320

Therefore, the total number of ways the "suites" can be arranged is given by:

2!\times9!\times8!=2\times362,880\times40,320=29,262,643,200 \ ways
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3 years ago
Of the entering class at a​ college, ​% attended public high​ school, ​% attended private high​ school, and ​% were home schoole
Veronika [31]

Answer:

(a) The probability that the student made the​ Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})

Compute the probability that the student made the​ Dean's list as follows:

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)

         =(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655

Thus, the probability that the student made the​ Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D)}

             =\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

P(C^{c}|D^{c})=1-P(C|D^{c})

               =1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

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Whats the question for this question 
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Please helps thank you
LUCKY_DIMON [66]
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Hope this helps:)
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