Nursing Assignment Help - Maths

Nursing Assignment Help Sample 

Assignment 2 Maths

Question 1 - There are 3 arrangements of the word DAD, namely DAD, ADD, and DDA. How many arrangements are there of the word PROBABILITY?

Arrangements in the word PROBABILITY

Formula: n! /a! b! c!....

Where n = total number of the terms and ‘a, b, c…’ are the different repeating terms.

In the term PROBABILITY, there are ‘2B’ and ‘2I’

Therefore, substituting in the formula, 11! /2! 2!

Number of arrangements

=11! ÷ 2!2!

=39,916,800 ÷ 4

Ans = 9,979,200

Question 2 - There are six men and seven women in a ballroom dancing class. If four men and four women are chosen and paired off, how many pairings are possible?

There are 6 men and 7 women

Selections: 4 men and 4 women for pairing

The total number of ways to select 4 out of the 6 men (6C4)

= 6!/(4!2!) = 15 quadruples

The total number of ways to select 4 out of the 7 women (7C4)

= 7!/(4!3!) = 35 quadruples

From the above two quadruples, any can select the respective partner, where the first man has 4 possible choices, second man has 3 choices, and the third one has 2 choices

Therefore, these pairings can occur in 4! Ways = 24 ways

Therefore, total number of unique pairs

=15 × 35 × 24

=12,600

Question 3 - Let A and B be two events. Suppose the probability that neither A or B occurs is 2/3. What is the probability that one or both occur?

Events: A and B

Probability that neither A or B occurs is 2/3

Probability that one or both occur means that A occurs or does not, B occurs or does not.

Therefore, the four combinations that sum up to 1 are presented as;

P (A ∩B) ∪ P (A ∩B^!) ∪P (A^! ∩B) ∪P (A^! ∩B^!) which sums to 1

We know that

P (A^! ∩B^!) = 2/3

Notably, the probability that one or both occur is;

P (A ∩B) ∪ P (A ∩B^!) ∪P (A^! ∩B)

This means;

P (A ∩B) + P (A ∩B^!) + P (A^! ∩B) is equivalent to

1 -  P (A^! ∩B^!)

= 1 - 2/3

Ans = 1/3

Question 4 - Let C and D be two events with P (C) = 0.25, P (D) = 0.45, and P (C ∩ D) = 0.1. What is P (Cc ∩ D)?

Events: C and D

We know that P(C) = 0.25, P(D) = 0.45, and P(C∩D) = 0.1

P(C∪D) = P(C) + P(D) – P(C∩D)

This means that P(C∪D) = 0.25 + 0.45 – 0.1

Therefore, P(C∪D) = 0.6

We do not know if C and D are independent. Therefore,

P(Cc∩D) = -P(C∩D) + P(D)

P(Cc∩D) = -0.1 + 0.45

Ans = 0.35

Question 5 - Suppose you are taking a multiple-choice test with c choices for each question. In answering a question on this test, the probability that you know the answer is p. If you don’t know the answer, you choose one at random. What is the probability that you knew the answer to a question, given that you answered it correctly?

There are c choices for every question

Probability of knowing the question is p

Random selection occurs if one does not know the answer;

If the student guesses the answer and is correct from the choices, the probability is 1/c

Let A be answering it correctly

Let B be knowing the answer, which is p

Therefore,

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Ans = p

Question 6 - A multiple-choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible.

4 choices of the multiple-choice exam.

Probability of knowing the answer is 0.5

Probability of eliminating a choice is 0.25 (otherwise all choices appear equally plausible)

Knowing the answer means getting it right

Let A represent the event where the question has been answered right

Let X be the event that the student knows the correct answer

Let Y be the event that the student is able to eliminate a choice

Let Z represent the event that the student is unable to eliminate any choice

Interpreting 0.5 as the probability of knowing the answer, 0.25 as the probability to be able to eliminate a choice, otherwise the 4 choices are equally plausible, this would mean,

P(X) = 0.5;P(Y) = 0.25;p(Z) = 0.25

Applying Baye’s Theorem

=24/31 or 77.4%

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