**I will answer this Question later**

- P(A)
- P(B)
- P(A intersect B)
- P(A U B) directly from the elementary events in the set and then using the addition rule:

P(A) + P(B) - P(A intersect B)

X | 1 | 2 | 3 | 4 | 5 |

P(X) | .6 | .2 | .1 | .07 | .03 |

Find the following:

- P(asking more than once) = 1 - .6 = .4
- P(asking more than twice given that the patient must ask more than once) =

P({3,4,5} intersect {2,3,4,5}, given that {2,3,4,5} occured)=

P({more than twice} intersect {more than once})/P(more than once) =

P( {3,4,5} intersect {2,3,4,5})/ P({2,3,4,5}) =

P({3,4,5})/P({2,3,4,5}) = .2/.4 = .5 - P(patients gets the pan within 3 requests) = P({1, 2, or 3})= .6 + .2 + .1 = .9
- P(asks less than twice or more than 4 times) P({1 or 5})= .6 + .03 = .63

x | [less than 3.5) | [3.5 to 4.5) | [4.5 to 5.5) | [5.5 to 6.5) | [6.5 and above |

P(X) | .15 | .23 | .4 | .17 | .05 |

Note: [a , b ) is read " [include in the class interval the numbers from "a" up to, but not including, the number "b")

Find:

- P(of a level between [3.5 and 5.5)) = .23 + .4 = .63
- P(of a level equal to or greater than 4.5) = .4 + .17 + .05 = .62
- P(of a level between [4.5 and 6.5)) = .4 + .17 = .57
- Using the Addition Rule find P({of levels between 3.5 and 5.5} or {greater than or equal to 4.5}) . Show your work.

A is the set {of levels between [3.5 to 5.5}

B is the set {greater than or equal to 4.5}

The intersection of A and B is the set { levels between [4.5 to 5.5}

P(A) = .63 P(B) = .62 P(A intersect B) =.4 P(A U B) = P({greater than or equal to 3.5}) = P(A) + P(B) - P(A intersect B) = .63 + .62 - .4 = .85 - P( levels less than 4.5) =.15 + .23 + .38

- P(of levels less than 3.5 given that the values are less than 6.5) = .15/.95 = 3/19
- P(of levels between 3.5 and 5.5 given than the values are 3.5 or greater) = .63 / .85
- P(of levels 6.5 or greater given that the values are 4.5 or greater) = .05/.62

**5. Three home care nurses have 25, 35, and 40 patients respectively. The severity is such that 5, 4, and 2 percent, respectively have to be rehospitalized within one month of discharge. One of their patients is readmitted to your unit. What is the probability that the patient came from the cases of nurse 1? nurse2? nurse3?**

- First we have to understand the question in terms of a question about probabilities. The question is asking: Given that a patient is readmitted, what is the probability that the patient was cared for by nurse A.? B? C?... This tells us we need the joint distribution of nurses and readmission status since the values needed to compute conditional probabilities, P(A intersect B) / P(B), are contianed in the joint distribution and its marginal distributions. So what joint distribution are we looking for? Let's sketch it out and fill in what we know

readmitted not readmitted A ? ? .25 B ? ? .35 C ? ? .40 ? ? 1.0 - To solve this problem (Given that a patient is readmitted, what is the probability that the patient was cared for by nurse A.? B? C?..) ... we have to find the marginal probability of being readmitted and also the joint probabilities that a patient was readmitted and had a particular nurse. These are not given directly. So what do we know?

The stem tells us the following:There are two variables of interest. The sample space for the first variable is ({readmitted} {not readmitted}) and the sample space for the other vairable is nurse ({A,} {B} {C}). - The problem stem also tells us ...
- the probability of being admitted given that you are a patient of nurse A = .05,
- the probability of being admitted given that you are a patient of nurse B =.04
- the probability of being admitted given that you are a patient of nurse C =.02
- it also tells us the number (or frequency) of patients cared for by nurse A, nurse B and nurse C. We also know there are 100 patients total (25 + 35 +40 = 100) From this we can find out the relative freuency (or probability) of being cared for by each of the nurses.

The marginal probability for the variable "nurses" are

A B C 25/100 =.25 35/100 =.35 40/100 = .40

- from our knowledge of probability we know
- P(being admitted given "say"... nurse A) = P(admiited and nurse A)/P(cared for by nurse).

- Using this information and the marginal probability for "nurses", we can find the joint probabilities we need in our table: P(admiited and nurse A), P(admiited and nurse B). P(admiited and nurse C).
- P(being admitted given "say"... nurse A) = P(admitted and nurse A)/P(cared for by nurse A) says

" .05 = P(admitted and nurse A)/.25".

We can solve for P(admitted and nurse A) using simple algebra

P(admiited and nurse A) = .05 * .25 = .0125

Similarly,

- P(admitted and nurse B) = .04 * .35 = .014
- P(admitted and nurse C) = .02 * .4 = .008
Let's put these in our joint probability table and use them to fill out the table as we can....:

readmitted not readmitted A .0125 (.25 - .0125) .25 B .014 (.35 - .014) .35 C .008 (.40 - .008) .40 .0345 (1 - .0345) 1.0

- P(being admitted given "say"... nurse A) = P(admitted and nurse A)/P(cared for by nurse A) says
- To find probability of being admitted we need to add up the following joint elementary events
{ (admitted and nurse A) (admitted and nurse B) (admitted and nurse C).< BR> .0125 + .014 + .008 = .0345

- P(cared for by nurse A given readmitted) = P(cared for by nurse A and readmitted)/P(of readmission ) = .0125/.0345 = .3623
- P(cared for by nurse B given readmitted) = P(cared for by nurse B and readmitted)/P(of readmission ) = .014/.0345 = .4058
- P(cared for by nurse C given readmitted) = P(cared for by nurse C and readmitted)/P(of readmission ) = .008/.0345 = .2319

This is the conditional distribution of nurse giving care given (or among) patients who were readmitted. Let's look at it expressed as a condititional probability distribution in a table.##### The Codnitional Distribution of Nurse given Readmits

A .3623 B .4058 C .2319

Now we are ready to answer the questions posed in the problem:

**6. A nurse wants to look at the effect of diagnosis on dietary intake of fluids. She observes patients with diseases A, B, and C and records the following intakes during waking hours. The probability distribution is as follows.**

A | B | C | |

400 or less | .20 | .04 | .10 |

400 to 800 | .05 | .15 | .15 |

more than 800 | .01 | .05 | .25 |

- What is the marginal probability for intake during the waking hours. Hint: how is the group distributed on intake?
A B C 400 or less .20 .04 .10 .34 400 to 800 .05 .15 .15 .35 more than 800 .01 .05 .25 .31 .26 .24 .50 1.0 - What is the marginal distribution of diseases.
**See above.** - What are the elementary events in the joint sample space.

{(400 A) (400 B) (400 C) (800 A) (800 B) (800 C) (more A) (more B) (more C)} - What is the conditional probability distribution for intake given disease A? given disease B? given disease C?

Given A Given B Given C 400 or less .20/.26 = .769 .04/.24 = .1667 .10/.50 = .2 400 to 800 .05/.26 = .19 .15/.24 = .625 .15/.50 = .3 more than 800 .01/.26 = .385 .05/.24 = .208 .25/.50 = .5 1.0 1.0 1.0 - Looking at these three conditional distributions, is there evidence for independence of the variables: disease and intake.
WEe use the definition of independence: P(A/B) = P(A) if A and B are independent.

The conditionals for intake in disease A, B and C are not all equal to the marginal distribution of intake. Therefore the variables disease and intake are dependent. - Find the probability that patients have (800 or less ) or (diseases B or C). Uses the addition rule. (Remember that you have to figure out what elementary events are in the intersection of these events and then knwo or be able to figure out probability of the intersection of them to use the addition rule.) After using the addition rule, write down the elementary events of interest and find the probability directly.

Using Addition Rule: .69 + .74 - .44 = .99 Directly: .20 + .05 + .04 + .15 + .05 + .25 +.10 +.15 = .99 - Use the definition of independence, P(A intersect B) = P(A)P(B) which says "for any two events,

the P(A intersect B) is equal to P(A) times P(B) if A is independent of B", to test whether having an intake of more than 800 cc is independent of disease category (A? B? C?).- P(+800 and A) = .01 is not equal to P(+800) * P(A) = .31 * .26 = .08 so having intake more than 800 is related to disease A
- P(+800 and B) = .05 is not equal to P(+800) * P(B) = .31 * .24 = .074 so having intake more than 800 is related to disease B
- P(+800 and C) = .25 is not equal to P(+800) * P(C) = .31 * .50 = .155 so having intake more than 800 is related to disease C

need dependent care | no dependent care | ||

marital problems | .20 | .133 | .33 |

no marital problems | .40 | .267 | .66 |

.60 | .40 | 1.0 |