1. 1. Basics of R 2. Basic probability with R
CA200
Dublin City University
(based on the book by Prof. Jane M. Horgan)

2. Installing R Go to the Cran website at http://cran.r-project.org/
Click ‘Download and Install R ’
Choose an operating system e.g. Windows;
Choose the ‘base’ package
Select the setup program (e.g. R *.exe)
Press the option ‘Run’
R is now installed :)
CA200

3. R Documentation - Manuals
An Introduction to R
The R Language Definition
Writing R Extensions
R Data Import/Export
R Installation and Administration.
R Internals
The R Reference Index
CA200

4. Basics 6+7*3/2 #general expression [1] 16.5
x <- 1:4 #integers are assigned to the vector x
x #print x
[1]
x2 <- x**2 #square the element, or x2<-x^2 x2
[1]
X < #case sensitive!
prod1 <- X*x
prod1
[1]
CA200

5. Basics <- assignment operator
R is case sensitive - x and X are different values
Variables can consist of any combination of cases or numbers but cannot begin with _ or numeral
Objects: The entities that R creates and manipulates, e.g. variables, arrays, strings, functions
Workspace: All objects created in R are stored in a workspace.
CA200

6. Getting Help click the Help button on the toolbar help() help.start()
demo()
?read.table
help.search ("data.entry")
apropos (“boxplot”) - "boxplot", "boxplot.default", "boxplot.stat”
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7. Data Entry Entering data from the screen to a vector Example:
downtime <-c(0, 1, 2, 12, 12, 14, 18, 21, 21, 23, 24, 25, 28, 29, 30,30,30,33,36,44,45,47,51)
mean(downtime)
[1]
median(downtime)
[1] 25
range(downtime)
[1] 0 51
sd(downtime)
[1]
CA200

8. 1. Basics of R 2. Basic probability with R
CA200

9. Basic definitions
Experiment is a process of observation that leads to a single outcome that cannot be predicted with certainty
Examples:
1. Pull a card from a deck
2. Toss a coin
3. Response time
Sample Space: All outcomes of an experiment. Usually denoted by S
Event denoted by E is any subset of S
1. E = Spades
2. E = Head
3. E = Component is functioning
P(E) denotes the probability of the event E
1. P(E) = P(Spades)
2. P(E) = P(Head)
3. P(E) = P(Component is functioning)
CA200

10. Calculating Probabilities
CLASSICAL APPROACH:
Assumes all outcomes of the experiment are equally likely:
Example: Roll a fair die.
E = even number
RELATIVE FREQUENCY APPROACH:
Interprets the probability as the relative frequency of the event over a long series of experiment.
Example: Roll a die a large number of times and observe number of times an even number occurs

11. Permutations
The number of ordered sequences where repetition is not allowed, i.e. no element can
appear more than once in the sequence.
Ordered samples (sequences) of size k from n:
𝑛 𝑃 𝑘 =𝑛 𝑛−1 … 𝑛−𝑘+1 = 𝑛! 𝑛−𝑘 !
In R use function:
prod
Example 1. Three elements {1,2,3}. How many sequences of two elements from these three?
Example 1 - Solution:
(1,2); (1,3); (2,1); (2, 3); (3,1); (3,2)  Six ordered sequences altogether.
3P2 = 3 * 2 = 6
Solution is R:
prod (3:2)
[1] 6
CA200

12. Permutations - examples
Q2. Four elements {1,2,3,4}. How many sequences of two elements from these four? Q3. Four elements {1,2,3,4}. How many sequences of three elements from these four?
Q2-solution:
(1,2); (1,3); (1,4) (2,1); (2, 3); (2,4); (3,1); (3,2); (3,4); (4,1); (4,2); (4,3).
Twelve ordered sequences altogether.
4P2 = 4 * 3 = 12
In R:
prod (4:3)
[1] 12
Q3-solution:
(1,2,3); (1,3,2); (1,2,4); (1,4,2); (1,3,4); (1,4,3)
(2,1,3); (2,3,1); (2,1,4); (2,4,1); (2,3,4); (2,4,3)
(3,1,2); (3,2,1); (3,2,4); (3,4,2); (3,1,4); (3,4,1)
(4,1,3); (4,3,1); (4,1,2); (4,2,1); (4,3,2); (4,2,3)
Twenty-four ordered sequences.
4P3 = 4 *3 * 2= 24
prod (4:2)
[1] 24

13. Combinations
The number of unordered sets of distinct elements, i.e. repetition is not allowed.
Number of ways of selecting k distinct elements from n or equivalently number of
unordered samples of size k, without replacement from n
𝑛 𝐶 𝑘 = 𝑛 𝑃 𝑘 𝑘! = 𝑛! 𝑘! 𝑛−𝑘 !
In R use function:
choose
Example 4. Three elements {1,2,3}. How many sets (combinations) of two elements from these three?
Example 4 - Solution:
{1,2}; {1,3}; {2,3}  Three ordered sequences altogether.
3C2 = (3*2)/2*1 = 3
In R:
choose (3,2)
[1] 3

14. Combinations - examples
Q5. Four elements {1,2,3,4}. How many combinations of two elements from these four?
Q6. Four elements {1,2,3,4}. How many unordered of three elements from these four?
Q7. If a box contains 75 good IC chips and 25 defective chips, and 12 chips are selected at random,
find the probability that all chips are good.
Q5-solution:
{1,2}; {1,3}; {1,4} {2,3}; {2, 4}; {3,4}  Six unordered sequences altogether.
4C2 = (4*3)/(2*1) = 6
In R: choose (4, 2)
[1] 6
Q6-solution:
{1,2,3}; {1,2,4}; {2,3,4}; {3,1,4}  Four unordered sequences.
4C3 = (4*3*2)/(3*2*1)= 4
In R: choose(4, 3)
[1] 4
Q7-Solution:
E – all chips are good 𝑃 𝐸 =
In R: 𝑃 𝐸 = 𝑐ℎ𝑜𝑜𝑠𝑒(75, 12) 𝑐ℎ𝑜𝑜𝑠𝑒(100, 12)

15. Bayes’ rule - example
Q8. In a computer installation, 200 programs are written each week, 120 in
C++ and 80 in Java. 60% of the programs written in C++ compile on the first
run and 80% of the Java programs compile on the first run.
If a randomly selected program compiles on the 1st run what is the probability that it was written in C++?
Q8 – Solution:
Lets denote:
C++ - event that program is written in C++
J - event that program is written in Java
E - event that selected program compiles on the first run
P(E) = P(C++)P(E|C++) + P(J)P(E|J) = 120/200 * 72/ /200*64/80 = 0.68
In R:
total_prob <- ((80/200)*(64/80)) + ((120/200)*(72/120)) #total probability
condit_c <- (120/200)*(72/120)/total_prob #posterior probability of C++
condit_c
[1]

16. Solve following exercises using R: (E2
Solve following exercises using R: (E2.1) A card is drawn from a shuffled pack of 52 cards. What is the probability of drawing a ten or a spade? (E2.2) Records of service requests at a garage and their probabilities are as follows: Daily demand is independent (e.g. tomorrow’s demand is independent of today’s demand). What is the probability that over a two-day period the number of requests will be a) 10 requests, b) 11 requests and c) 12 requests? (E2.3) Analysis of a questionnaire completed by holiday makers showed that 0.75 classified their holiday as good at resort Costa Lotta. The probability of hot weather in this resort is 0.6. If the probability of regarding the holiday as good given hot weather is 0.9, what is the probability that there was hot weather if a holiday maker considers his holiday good?
Daily Demand
Probability 5
0.3 6
0.7
CA200

17. p(ten) = 4/52, p(spade) = 13/52 = ¼, p(ten of spades) = 1/52.
Solutions:
E2.1 A card is drawn from a shuffled pack of 52 cards. What is the probability of drawing a ten or a spade?
E2.1-Solution:
p(ten) = 4/52, p(spade) = 13/52 = ¼, p(ten of spades) = 1/52.
Therefore, p(drawing a ten or a spade) = 4/ /52 – 1/52 = 16/52 = 4/13.
E2.2 Records of service requests at a garage and their probabilities are as follows:
Daily demand is independent (e.g. tomorrow’s demand is independent of today’s demand).
What is the probability that over a two-day period the number of requests will be
10 requests, b) 11 requests and c) 12 requests?
E2.2 - Solution:
a) p(10 requests) = p(5 and 5) = 0.3*0.3 = 0.09
b) p(11 requests) = p(5 and 6) + p(6 and 5) = 0.3* *0.3 = 0.42
c) p(12 requests) = p(6 and 6) = 0.7*0.7 = 0.49
E2.3 Analysis of a questionnaire completed by holiday makers showed that 0.75 classified their holiday as good at resort Costa Lotta. The probability of hot weather in this resort is 0.6. If the probability of regarding the holiday as good given hot weather is 0.9, what is the probability that there was hot weather if a holiday maker considers his holiday good?
E2.3 - Solution: Let H = hot weather (event), G = good holiday (event) => we need p(H|G).
We are given p(G) = 0.75, p(H) = 0.6 and p(G|H) = 0.9.
Bayes’ theorem => p(H|G) = p(H)p(G|H)/p(G) = 0.6*0.9/0.75 = 0.72 or 72% chance.
Daily Demand
Probability 5
0.3 6
0.7

18. Binomial distribution
Def:
Conditions:
An experiment consists of n repeated trials
Each trial has two possible outcomes
A success with probability p is constant from trial to trial
A failure with probability q = 1 − p
3. Repeated trials are independent
X = number of successes in n trials, X is a BINOMIAL RANDOM VARIABLE.
The probability of getting exactly x successes in n trials is:
In R use function:
use prefix “binom” with prefix “d” for pdf. Function is dbinom(x, n, p)

19. Binomial - examples
Q9. Five terminals on an on-line computer system are attached to a communication line to the central computer system. The probability that any terminal is ready to transmit is Q9-solution: Lets denote: X = number of ready terminals p – probability of success in one trial n – number of trials p = 0.95 n = 1 – q = 5 P(X=0) = P(0 ready terminal in 5): FFFFF P(X=1) = P(1 ready terminal in 5): SFFFF, FSFFF, FFSFF, FFFSF, FFFFS In R: For all cumulative probabilities: x <- 0:5 dbinom(x, size = 5, prob = 0.95) [1] The probability of getting exactly 3 ready terminals from five: dbinom(x = 3, size =5, prob = .95) [1]

20. Binomial - examples
Q9. Five terminals on an on-line computer system are attached to a communication line to the central computer system. The probability that any terminal is ready to transmit is 0.95.
Q9-solution – using plot function
plot(x, dbinom(x, size = 5, prob = 0.95), xlab = "Number of Ready Terminals",
ylab = "P(X = x)", type = "h", main = "Ready Terminals, n = 5, p = 0.95")

21. Binomial - examples
Q10. A fair coin is tossed 10 times. Success and failure are “heads” and “tails”, respectively, each with probability, 0.5. Q10-solution: X = number of heads (successes) obtained p = 0.50 n = 10 P(X=0) = P(exactly 0 head in 10 tosses) P(X=1) = P(exactly 1 head in 10 tosses) P(X=2) = P(exactly 2 heads in 10 tosses)… In R: x <- 0:10 round(dbinom(x, 10, 0.5), 4) [1]

22. Binomial - examples
Q10. A fair coin is tossed 10 times. Success and failure are “heads” and “tails”, respectively, each with probability, 0.5. Q10 solution – using plot function plot(x, dbinom(x, size = 10, prob = 0.5), xlab = "Number of Heads", ylab = "P(X = x)", type = "h", main = "Heads, n = 10, p = 0.5")