ISP Notes

Limitations of Probability

When the number of outcomes of the sample space is infinite, it cannot be used to find the probability of an event.
If the total number of outcomes of the sample space is unknown, the probability of an event cannot be obtained.
If all possible outcomes of the sample space are not equiprobable, the probability of an event cannot be obtained.

Exhaustive - The events that make up the sample space
Mutually exclusive - Two events that cannot occur simultaneously
Axiomatic definition of probability
There are some events where getting exact denomator is not possible (infinite)

1) Two dices are thrown together. What is the probability of getting a sum of 12?
A: 1/36
2) Two cards are drawn at random from a full pack of 52 cards. Find the probability that (i) Both are red (ii) One is a heart and other is a diamond
A:
(i) 26/52 * 25/51 = 25/102
(ii) 13/52 * 13/51 = 13/204
3) If 10 persons are arranged at random in (i) a line (ii) a circle, find the probability that any two persons will be next to each other.
A:
(i) (9! * 2!)/10! = 1/5
(ii) (8! * 2!)/9! = 2/9
4) An urn contains 8 white and 3 red balls. If 2 balls drawn are drawn at random, find the probability that (i) both are white (ii) both are red (iii) one of each colour
A:
(i) 8C2/11C2 = 28/55
(ii) 3C2/11C2 = 3/55
(iii) (8C1 * 3C1)/11C2 = 24/55
5) 4 cards are drawn at random from a full pack. What is the probability they belong to (i) Four different suites (ii) Different suites and denominations.
A:
(i) (13C1)^4/52C4 = 2197/20825
(ii) (52 * 36 * 22 * 10)/(52 * 51 * 50 * 49) = 264/4165
6) What is the probability that all three children in a family have different birthdays?
A: (365 * 364 * 363)/(365 * 365 * 365) = 132132/133225 = 99.18%
7) What is the probability that a leap year has 53 Mondays?
A: 2/7
8) A bag contains 8 white and 6 black balls. If 5 balls are drawn at random, find the probability that 3 are white and 2 are black.
A: (8C3 * 6C2)/14C5 = 60/143
9) 5 men in a company of 20 are graduates. If three are picked at random, find the probability that (i) all are graduates (ii) at least one is a graduate
A:
(i) 5C3/20C3 = 1/114
(ii) (5C1 * 15C2 + 5C2 * 15C1 + 5C3)/20C3 = 137/228
10) In a single throw of two dice, find the chance of getting a sum of 7
A: n(Sample space) = 6^2 = 36
Sample space for sum of 7 = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} => n(Sample space for sum of 7) = 6
=> Chance of getting 7 = 6/36 = 1/6
11) A four digit number is formed by the digits 1,2,3,4 with no repetition. Find the probability that the number is (i) odd (ii) divisible by 4
A:
(i) (3! * 2)/4! = 1/2
(ii) (2! * 3)/4! = 1/4
12) There are four persons in a company. Find the probability that (i) All have different birthdays (ii) At least two have same birthdays (iii) Exactly two have same birthday
A:
(i) (365 * 364 * 363 * 362)/365^4 = 47831784/48627125
(ii) (365 * 1 * 364 * 363)/365^4 + (365 * 1 * 1 * 364)/365^4 + (365 * 1 * 1 * 1)/365^4 = 132497/48627125
(iii) (365 * 1 * 364 * 363)/365^4 = 132132/48627125**

P(A+B)=P(A)+P(B) if mutually exclusive
P(A+B)=P(A)+P(B)-P(AB) if not
**P(A+B+C)=P(A)+P(B)-P(AB)-P(BC)-P(AC)+P(ABC)
P(A1 ∪ A2 ∪ · · · ∪ An) = https://personal.utdallas.edu/~metin/Probability/MyNotes/independenceBayes.pdf
Boole's inequality:**
P(A+B)<=P(A)+P(B)
Bonferroni's inequality:
P(AB)>=P(A)+P(B)-1
Theorem of compound probability:
P(AB)=P(A)P(B|A)
P(B|A)=P(AB)/P(A)
P(ABC)=P(A)P(B|A)P(C|(AB))

27/2/23
13) P(A)=1/2, P(B)=1/3, P(AB)=1/4. Find i) P(A') ii) P(A+B) iii) P(A|B) iv) P(A'B) v) P(A'B') vi) P(A'+B)
Answer:
i) P(A') = 1-P(A) = 1 - 1/2 = 1/2
ii) P(A+B) = P(A)+P(B)-P(AB) = 1/2 + 1/3 - 1/4 = 7/12
iii) P(A|B) = P(AB)/P(B) = (1/4) / (1/3) = 3/4
iv) P(A'B) = P(B) - P(AB) = 1/3 - 1/4 = 1/12
v) P(A'B') = P((A+B)') = 1 - P(A+B) = 5/12
vi) P(A'+B) = P(A')+P(B)-P(A'B) = 1/2 + 1/3 - 1/12 = 3/4
14) P(A) = 3/8, P(B) = 5/8, P(A+B) = 3/4. Find i) P(A|B) ii) P(B|A). Also state whether A and B are independent events.
Answer:
3/4 = 3/8 + 5/8 - P(AB) => P(AB) = 1/4
i) P(A|B) = P(AB)/P(B) = (1/4) / (5/8) = 2/5
ii) P(B|A) = P(AB)/P(A) = (1/4) / (3/8) = 2/3
P(AB) = 1/4 and P(A)P(B) = (3/8) * (5/8) = 15/64. Since P(AB)!=P(A)P(B), A & are not independent.
15) An article manufactured by a company consists of two parts I and II. In this process of manufacture of I, 9/100 are likely to be defective. 5/100 are likely to be defective in the manufacture of II. Calculate the probability that the assembled item will not be defective.
Answer:
P(A) = I not defective
P(B) = II not defective
We need to find P(AB)
Since A & B are independent events therefore P(AB)=P(A)P(B) = (91/100) * (95/100) = 1729/2000 is the probability that the assembled item is not defective.
16) A salesman has 80% chance of making a sale to each customer. The behaviour of successive customers is assumed to be independent. If two customers X & Y enter the shop, find the probability that the salesman will make a sale.
Answer:
P(X) = 8/10
P(Y) = 8/10
We need to find P(X+Y)
P(X+Y) = 1-P(X'Y')
Since X & Y are independent, P(X'Y') = P(X')P(Y')
=>P(X+Y)=1-P(X')P(Y') = 1 - (2/10) * (2/10) = 0.96.
Therefore probability that he makes a sale is 96%.
17) Two players A & B toss a die alternately. He who first throws a six wins the game. If A begins, what is the probability that (i) A wins the game (ii) B wins the game.
Answer:
i)
A wins if:
E1 = A throws a 6
E2 = A doesn't throw 6, B doesn't throw 6, A throws 6
E3 = A doesn't throw 6, B doesn't throw 6, A doesn't throw 6, B doesn't throw 6, A throws 6
...
En
We can see this is a geometric progression with a=1/6 and r=(5/6)^2
Therefore P(A wins) = a/(1-r) = (1/6) / (1-(5/6)^2) = 6/11
ii) P(B wins) = 1-P(A wins) = 5/11**

28/2/23
Binomial probability:

p = probability of success
q = 1-p (probability of failure
r = no. of success (success is defined)
n = total number of trials
P(r) = nCr * p^r * q^(n-r)
18 The probability that an entering college student will be a graduate is 0.4. Determine the probability that out of 5 entering students i) none will graduate ii) one will graduate iii) at least one will graduate
Answer:
p = 0.4, q = 0.6, n=5
i)
r=0
P(none will graduate) = 5C0 * 0.4^0 * 0.6^(5-0) = 0.07776
ii)
r=1
P(one will graduate) = 5C1 * 0.4^1 * 0.6^(5-1) = 0.2592
iii)
r=1,2,3,4,5
P(at least one will graduate) = 1 - P(none will graduate) = 1 - 0.07776 = 0.92224
19) A machine produces on average 20% defects. If 4 articles are chosen randomly, find the probability that there are at least 2 defective articles.
p=0.2, q=0.8, n=4, r=2,3,4
P(at least 2 are defective) = (4C2 * 0.2^2 * 0.8^2) + (4C3 * 0.2^3 * 0.8^1) + (4C4 * 0.2^4 * 0.8^0) = (6 * 0.04 * 0.64) + (4 * 0.2^3 * 0.8^1) + (1 * 0.0016 * 1) = 0.1536 + 0.0256 + 0.0016 = 0.1808 = 18.08% probability that at least two of the four articles are defective.

Bayes theorem:
An event A can occur only if one of the mutually exclusive and exhaustive set of events B1, B2, ... , Bn occurs.
Suppose that the unconditional probabilities P(B1), P(B2), ... , P(Bn) and conditional probabilities P(A|B1), P(A|B2), ... , P(A|Bn) are known. Then the conditional probability P(Bi|A) of a specified event B, when A is stated to have actually occurred is given by:
P(Bi|A) = (P(Bi) * P(A|Bi)) / Σ (P(Bi) P(A|Bi))
where Σ = i from 1 to n.
20) Two boxes contain respectively 4 white and 2 black balls, 1 white and 3 black balls. One ball is transferred from first box to the second. Then a ball is drawn from the second box. It turns out to be black. What is the probability that the transferred ball was white?
Answer:
Let B1 = transferred ball is white
B2 = transferred ball is black
A = black ball is drawn from second box
=> P(B1) = 4/6
=> P(B2) = 2/6
P(A|B1) = 3/5
P(A|B2) = 4/5
Therefore P(B1|A) = ((4/6) * (3/5)) / ((4/6) * (3/5) + (2/6) * (4/5)) = 3/5
21) A doctor is called to see a sick child. The doctor has prior information that 90% of sick children in that neighborhood have flu, while the other 10% are sick with measles. Let F stand for an event of child being sick with flu and M stand for an event of a child being sick with measles. A well-known symptom of measles is a rash (the event of having which we denote as R). Assume that the probability of having a rash if one has measles is P(R|M)=0.95 and P(R|F)=0.08. Upon examining the child the doctor finds a rash. Find the probability that the child has measles.
Answer:
P(M|R) = (P(M)P(R|M)) / (P(M)P(R|M) + P(F)P(R|F))
=> (0.1 * 0.95)/(0.1 * 0.95 + 0.9 * 0.8) = 0.095/0.167 = 95/167 = 0.568

22) Three boxes of same type are there with following number of white and black balls as follows - Box I: 1 white, 2 black, Box II: 2 white & 1 black, Box III: 2 white & 2 black. One of the boxes is selected at random and one ball is drawn randomly from it. It turns out to be white. What is the probability that the third box is chosen?
Answer:
P(B3|W) = (P(B3)P(W|B3)) / (P(B1)P(W|B1)+P(B2)P(W|B2)+P(B3)P(W|B3)) = (1/3 * 1/2) / (1/3 * 1/3 + 1/3 * 2/3 + 1/3 * 1/2) = 1/3

23) In a group of 20 males and 5 females, 10 males and 3 females are service holders. What is the probability that a person selected at random is a service holder given that he is male?
Answer:
P(S) = 13/25
P(M)=20/25
P(S|M)= P(SM)/P(M) = (10/25)/(20/25) = 1/2
24) A picnic is arranged to be held on a particular day. The weather forecast says that there is 80% chance of rain on that day. If it rains, the probability of a good picnic is 0.3, and if it does not, probability is 0.9. What is the probability that the picnic will be good?
Answer:
P(R) = 0.8
P(R')=0.2
P(G|R)=0.3
P(G|R')=0.9
P(G)=P(R)P(G|R)+P(R')P(G|R') = 0.8 * 0.3 + 0.2 * 0.9 = 0.42
25) The probability that X, Y and Z become principal of a certain college are 0.3, 0.5 and 0.2 respectively. The probability that a 'student aid fund' will be introduced in the college if X, Y and Z become principal are 0.4, 0.6 and 0.1 respectively. Given that the student aid fund is introduced, find the probability that Y has been appointed.
P(Y|A) = (P(Y)P(A|Y)) / (P(X)P(A|X) + P(Y)P(A|Y) + P(Z)P(A|Z)) = (0.5 * 0.6)/(0.3 * 0.4 + 0.5 * 0.6 + 0.2 * 0.1) = 0.3/(0.12+0.3+0.02) = 0.3/0.44 = 15/22
26) In a certain city, suppose that 3% of the population is known to be affected by a particular disease. There is a test for the disease. Of those with the disease, 98% test positive, and those without the disease, 99.8% test negative. What is the probability that an individual selected at random with a positive result does not have the disease?
P(D) = 0.03 -> P(D') = 0.97
P(T|D) = 0.98
P(T|D') = 0.002
=> P(D'|T) = (P(D')P(T|D'))/(P(D)P(T|D)+P(D')P(T|D')) = (0.97 * 0.002)/(0.97 * 0.002 + 0.03 * 0.98) = 0.00194/0.03134 = 0.0619 = 97/1567
27) What is the probability that a permutation of the letters 'MISSISSIPPI' result in all the 4S being together?
M = 1
I = 4
S = 4
P = 2
Total possible permutations = (4+4+2+1)! / (4! * 4! * 2! * 1!) = 11! / (4! * 4! * 2! * 1!) = 34650
Since 4 S need to appear consecutively, they can be denoted as a single unit: M = 1, I = 4, S = 1, P = 2
Total permutations with S together = (4+1+1+2)!/(4! * 2!) = 840
=> Probability that 4S appear together consecutively = 840/34650 = 4/165

ISPU2&3

Sample space = collection of some outcomes (May or may not be numerical values)
Assigning a value to some outcome = random variable
Suppose three coins are tossed at once. The sample space is:
S={HHH, HHT, HTT, HTH, TTH, THT, THH, TTT}
Let X be the probability of getting x number of heads.
When x =0, P(X=0) = 1/8
x=1, P(X=1) = 3/8
x=2, P(X=2) = 3/8
x=3, P(X=3) = 3/8
X = random variable (in this case number of heads).
Random variable is a real value function defined on the sample space for a random experiment.
In the above case, it is discrete.
f(x) = Probability mass function (for discrete).
A probability mass function (pmf) is a function over the sample space of a discrete random variable X which gives the probability that X is equal to a certain value.
f(x) must satisfy two conditions:

f(x) >= 0 ∀ x ∈ S
Σ f(x) = 1 (summation of x ∈ S)

Continuous Probability Distribution

f(x) = 2x when 0<x<=1, 4-2x when 1<x<=2 (i)
Suppose x = 0.253.
=> f(x) = P(X=x)
=> P(X=0.253)
=> 2 * 0.253 = 0.506
If x = 1.313
=> f(x) = P(X=x)
=> P(X=1.313)
=> 4 - 2 * 1.313 = 1.374
Therefore here f(x) = Probability density function (for continuous)
Conditions:

f(x) >= 0
∫ f(x)dx = 1 (integration running from -∞ to ∞)
For P(c<=x<=d)
= ∫ f(x)dx (integration running from c to d).
Let c=0.2, d = 0.7 then f(x) = ∫ f(x)dx (integration running from 0.2 to 0.7).
f(x) = 2x (from (i))
=> ∫ 2xdx (integration running from 0.2 to 0.7).
=> 2∫xdx (integration running from 0.2 to 0.7)
=> 2(x^2/2) (x running from 0.2 to 0.7)
=> x^2 (x running from 0.2 to 0.7)
=> 0.7^2 - 0.2^2 = 0.45
Therefore P(c<=x<=d) = 0.45 = 9/20
Now let's check whether (i) satisfies pdf.
We need to prove that ∫ f(x)dx = 1 (integration running from -∞ to ∞)
=>∫ 0dx (integration running from -∞ to 0) + ∫ 2xdx (integration running from 0 to 1) + ∫ (4-2x)dx (integration running from 1 to 2) + ∫ 0dx (integration running from 2 to ∞)
=>∫ 2xdx (integration running from 0 to 1) + ∫ (4-2x)dx (integration running from 1 to 2)
=> 2∫xdx (integration running from 0 to 1) + (∫4dx - ∫2xdx (integration running from 1 to 2))
=> x^2 (running from 0 to 1) + (4x - x^2 (running from 1 to 2))
=> (1^2 - 0^2) + ((4 * 2 - 2^2) - (4 * 1 - 1^2))
=> 1 + ((8-4) - (4-1))
=> 1+4-3 = 2
As ∫ f(x)dx != 1, therefore it does not satisfy PDF.

Sums

1) f(x) = 1/4 when -2<=x<=2, 0 for all other values. Prove that it is PDF.
A: ∫ (1/4)dx (integration running from -2 to 2)
=> x/4 (running from -2 to 2)
=> 2/4 - (-(2/4))
=> 1/2 + 1/2 = 1.
Therefore it satisfies PDF.

2) f(x) = cx^2 when 0<=x<=1, and 0 otherwise. f(x) is a PDF. Find i) c ii) P(0<=X<=1/2) iii) P(X>3/4)
A: Since f(x) is a PDF, therefore (i) ∫ cx^2dx = 1 (integration running from 0 to 1)
=> c∫x^2 dx = 1 (integration running from 0 to 1)
=> c(x^3/3) = 1 (running from 0 to 1)
=> cx^3 = 1 (running from 0 to 1)
=> c(1^3/3 - 0^3/3) = 1
=> c/3 = 1 => c=3
(ii) P(0<=X<=1/2)
=> ∫3x^2 (integration running from 0 to 1/2)
=> 3∫x^2 (integration running from 0 to 1/2)
=> 3(x^3/3) (running from 0 to 1/2)
=> 3((1/2)^3/3 - 0^3/3) = 1/8
(iii) P(X>3/4)
=>∫3x^2 (integration running from 3/4 to 1)
=> 3∫x^2 (integration running from 3/4 to 1)
=> 3(x^3/3) (running from 3/4 to 1)
=> 3(1^3/3 - (3/4)^3/3)
=> 3(1/3 - 9/64) = 37/64

3) f(x) = x when 0<=x<=1, k-x when 1<=x<=2, 0 otherwise. f(x) is a PDF. Find k.
A: x dx (integration running from 0 to 1) + (k-x)dx (integration running from 1 to 2) = 1
=> x^2/2 (running from 0 to 1) + (kx - x^2/2) (running from 1 to 2) = 1
=> 1/2 - k - 3/2 = 1
=> k=2

4) f(x) = 1/2 - ax, when 0<=x<=4, 0 otherwise. Find a. Find (i) P(X>=1) (ii) P(X>=2.5) (iii) P(|X-2|<0.5)

A: => ∫(1/2 - ax) dx (integration running from 0 to 4)
=> ∫1/2 dx - ∫axdx
=> [x/2 - ax^2/2] (running from 0 to 4)
=> 2-8a=1 => a=1/8

(i) P(X>=1)
∫(1/2 - 1/8x) dx (integration running from 1 to 4)
= 9/16

(ii) P(X>=2.5)
∫(1/2 - 1/8x) dx (integration running from 2.5 to 4)
= 9/64

(iii) P(|X-2|<0.5)
=> P(-0.5<X-2<0.5)
=> P(1.5<X<2.5)
=> ∫(1/2 - 1/8x) dx (integration running from 1.5 to 2.5)
= 1/4

Cumulative Distributive Function (CDF)

x	f(x)
0	1/8
1	3/8
2	3/8
3	1/8

=> F(x)=P(X<=x)

=> F(0) = P(X<=0) = P(X=0) = 1/8
=> F(1) = P(X<=1) = P(X=0) + P(X=1) = 1/2
=> F(2) = P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 7/8
=> F(3) = P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1

=> F(x) = 0 when x<0, 1/8 when 0<=x<1, 1/2 when 1<=x<2, 7/8 when 2<=x<3, 1 when x>=3.

Properties of F(x)

0<=F(x)<=1
P(c<=x<=d) = F(d) - F(c)
f(x) = d/dx F(x)

Expectation E(x):
i) Discrete: E(x) = Σxp(x)
ii) Continuous: E(x) = ∫xf(x)dx

No. of heads (x)	P(x)	xP(x)	(x-µ)^2 P(x)
0	1/8	0	9/32
1	3/8	3/8	3/32
2	3/8	3/4	3/16
3	1/8	3/8	27/32
		Σ = 3/2	Σ = 45/32

E(x) = 1.5
Var(x) = 1.40625
SD(x) = 1.18585

Binomial distribution

It is a PMF. (Discrete)

f(x) = nCx * p^x * q^(n-x)

E(x) = np
Var(x) = npq

1) Suppose 1/3 of the diabetic patients of a city are on insulin. Each of 100 investigators collect a sample of 10 individuals from population of diabetic patients to see if they are on insulin. How many of the investigators are expected to report that at most 3 out of 10 diabetic patients are on insulin?

A: p = 1/3, q=2/3, n=10
=> P(X<=3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)
=> 1024/59049+5120/59049+1280/6561+5120/19683 = 0.559

Therefore, investigators expected to report that at most 3 out of 10 diabetic patients are on insulin = 0.56 * 10 = 56.

2) Mean and SD of binomial distribution are 4 and √(8/3) respectively. Find p and q.

A: np=4
√(npq) = √(8/3)

=> 4q = 8/3
=> q = 2/3 => p=1/3

=> n=12

Poisson Distribution

Discrete distribution

X~P(λ)
λ = parameter

f(x) = (e^-λ * λ^x)/x!, x=0,1,2,...,n

Can be used for count per unit time

E(x) = λ = Var(x) = np

Skewness (γ) = 1/√λ
Kurtosis (γ2) = 1/λ

Poisson approximation to binomial distribution: If n is large and p is very small, we take np = λ.

1) The probability that a transistor manufacturing firm manufactures defective is 0.007. What is the probability that there is no defective in a batch of 50?

A: n=50, p = 0.007
=> λ = np = 0.35

=> P(X=0) = (e^-0.35 * 0.35^0)/0! = 0.70
=> 70% chance none are defective.

2) A hospital switchboard receives an average of 0.95 calls per minute. Find the probability that there are 2 or more calls in a minute. Also find average no. of calls per minute.

A: λ = 0.95
=> P(X>=2) = 1-P(X<2) = 1-[e^-0.95 * 0.95^0/0! + e^-0.95 * 0.95^1/1!]
=> 1-[0.39+0.37] = 0.245

3) In turning out certain toys in a manufacturing process in a factory, the average number of defectives is 10%. Find the probability of getting exactly 3 defectives in a sample of 10 toys chosen at random.

A: p = 10%, n = 10
λ = np = 1

=> P(X=3) = (e^-1 * 1^3)/3! = 0.06

4) A business firm receives on an average 2.5 telephone calls per day during the time period 10.00 - 10.05 a.m. Find the probability that on a certain day, the firm receives (i) no call; (ii) exactly 4 calls, during the same period. (Assume Poisson distribution; given e^-2.5 = .0821)

A: λ = 2.5

(i) P(X=0) = (e^-2.5 * 2.5^0)/0! = 0.0821
(ii) P(X=4) = (e^-2.5 * 2.5^4)/4! = 0.1336

5) Given X~P(λ) and P(X=1) = P(X=2). Find P(X<2).

A: P(X=1) = (e^-λ * λ^1)/1!
P(X=2) = (e^-λ * λ^2)/2!

Given (e^-λ * λ^1)/1! = (e^-λ * λ^2)/2!
=> 2λ = λ^2
=> λ = 2, λ ≠ 0.

=> P(X<2) = P(X=0)+P(X=1)
=> (e^-2 * 2^0)/0! + (e^-2 * 2^1)/2! = 0.406

E(x) = 2 = Var(x)

6) 2% of the items made by a machine are defective. Find the probability that 3 or more items are defective in a sample of 100 items.

A: p = 2%, n = 100
=> λ = np = 2

P(X>=3) = 1-P(X<3)

=> 1-[P(X=0)+P(X+1)+P(X=2)]
=> 1-(e^-2 + 2e^-2 + 4e^-2)
=> 1 - 5e^-2
= 0.323

7) A hospital switch board receives an average of 0.9 calls per minute. Find the probability that there is (i) no call in a minute, (ii) two or more calls in a minute. What is the average number of calls per minute? What is the standard deviation of the number of calls received per minute?

A: λ = 0.9

(i) P(X=0) = e^-0.9 = 0.406

(ii) P(X>=2) = 1-P(X<2) = 1-[P(X=0)+P(X=1)] = 1-(e^-0.9+0.9e^-0.9) = 1-1.9e^-0.9 = 0.2275

E(x) = 0.9, std(x) = 0.95

Geometric Distribution

X~Geo(P)

x = Number of failures before first success

x = 0, 1, 2, ..., ∞

p = probability of success
q = probability of failure

f(x) = q^(x-1) * p (PMF)

Properties

E(x) = 1/p
Var(x) = q/p^2
SD(x) = q/√p

CDF (P(X≤x)): F(x) = 1-q^x

1) If a patient is waiting for a suitable blood donor and the probability that the selected donor will be a match is 0.2, find the expected number of donors who will be tested till a match is found including the matched donor.

A: E(x) = 1/p = 1/0.2 = 5

2) Suppose you're playing a game of darts. Probability of success is 0.4. What is the probability that you'll hit the bull's eye on the 3rd try?

A: f(x) = q^(x-1) * p = 0.6^2 * 0.4 = 0.144

3) Calculate the probability density of geometric distribution if p=0.42. Also find out mean and variance.

A: CDF = 1-0.58^x

E(x) = 2.38
Var(x) = 0.58/0.42^2 = 3.28978

Sums

1) SD of Poisson distribution is 2. Find probability that x=3.

A: λ = 4
=> P(X=3) = (e^-4 * 4^3)/3! = 0.195

2) If probability of defective bolt is 1/10, find mean and variance for the distribution of defective bolts in a total of 400.

A: Mean = np = 1/10 * 400 = 40
Variance = npq = 1/10 * 400 * 9/10 = 36

3) Number of deaths per day in a city due to road accidents and due to other causes independently follow Poisson distribution with parameters 2 and 6 respectively. Find probability that total no. of deaths on a particular day is 2 or less.

A: λ1 = 2, λ2=6
=> λ = λ1+λ2 = 2+6 = 8
=> P(X≤2) = P(X=0)+P(X=1)+P(X=2) = 0.01375

4) Overall percentage of failures in a certain examination is 40. What is probability that out of 6 candidates, at least 4 passed the examination?

A: q=40%
p=60%
n=6

P(X≥4) = P(X=4)+P(X=5)+P(X=6) = 0.31+0.18+0.05 = 0.54

Normal Distribution

f(x) = 1/√(2π)σ * e^((-1/2)((x-µ)/σ)^2) [PDF]

Spread of curve directly depends on σ

µ, σ/σ^2 -> Parameters

X ~ Normal(µ,σ^2)

=> P(X>=1) = ∫ f(x) dx (integration from 1 to ∞) = ∫ 1/√(2π)σ * e^((-1/2)((x-µ)/σ)^2) dx (integration running from 1 to ∞)
=> f(x) = 1/√(2π)σ * e^((-1/2)(z^2)

Z~N(0,1)
z = Standard normal variate

To move from x to z, z=(x-µ)/σ

Total area under curve = Total probability = 1

-∞<=z<=∞

F(x) = ∫ f(x) dx (integration from -∞ to x) [CDF]

Properties

The normal distribution has two parameters µ & σ
Mean = µ, Standard deviation = σ
Mean = Median = Mode = µ
The quartiles are equidistant from mean.
Q1= µ - 0.67σ
Q3 = µ + 0.67σ
Quartile deviation = 0.67σ
Mean deviation = 0.80σ
All odd order central moments are zero.
µ1 = µ3 = 0
The second order and fourth order central moments are
µ2 = σ^2, µ4 = 3σ^4
In general, µ(2r+1) = 0 and µ(2r) = 1.3.5... (2r-1)σ^(2r)
β1 = 0, β2 = 3
Skewness (γ1) = 0, Kurtosis (γ2) = 0
The normal curve is bell-shaped and symmetrical about the line x=µ. The two tails of the curve extend to infinity on both sides of the mean. The maximum ordinate is at x=µ and given by
y = 1/(σ * √(2x))
The points of inflection of the normal curve are at x=µ±σ. This means that at these points the normal curve changes its curvature from concave to convex and vice-versa.
The percentage distribution of area under the 'standard normal curve' is:
Area between z=±1 is 68.27%
Area between z=±2 is 95.45%
Area between z=±3 is 99.73%
If a random variable x is normally distributed with mean µ and s.d. σ, then almost all the values of x will lie between the limits µ-3σ to µ+3σ, i.e., Mean±3SD.
If x and y are independent normal variates with means µ1 and µ2, and variances σ1^2 and σ2^2 respectively, then (x+y) is also a normal variate with mean (µ1+µ2) and variance (σ1^2+σ2^2).

Sums

1) The height distribution of a group of 10,000 men is normal with mean height = 64.5 and SD = 4.5. Find the number of men whose height is a) Less than 69 but greater than 55.5, b) Less than 55.5 c) More than 73.5.

A: x = height.

(i) P(55.5 < x < 69)
z = -2 when x = 55.5, z=1 when x = 69.
=> P(-2<z<1)
=> 95.45%/2 + 68.27%/2 = 81.86%
=> Number of men = 81.86% * 10,000 = 8,186
(ii) P(x<55.5)
P(z<-2)
=> 2.28%
=> Number of men = 2.14% * 10,000 = 228
(iii) P(X>73.5)
=> P(z>2)
=> 2.28%
=> Number of men = 2.14% * 10,000 = 228

2) The mean weight of 500 male students at a certain college is 151 lbs, and the standard deviation is 15 lbs. Assuming that the weights are normally distributed, find how many students weight (i) between 120 and 155 lbs., (ii) more than 155 lbs. [Given Φ(0.27) = 0.6064 and Φ(2.07) = 0.9808, where Φ(t) denotes the area under standard normal curve to the left of the ordinate at t.]

A: (i) P(120<=x<=155)
=> P(-2.07<=z<= 0.27) = Φ(0.27) - Φ(-2.07) = Φ(0.27) - (1- Φ(2.07)) = 0.6064 - 1 + 0.9808 = 0.5872
=> 58.72% * 500 = 294

(ii) P(X>155)
=> P(z>0.27) = 1 - P(z<=0.27) = 1 - Φ(0.27) = 1 - 0.6064 = 0.3936
=> 39.36% * 500 = 197

Uniform Distribution

f(x) = 1/(b-a) , a<=x<=b

P(c<=x<=d) = (d-c)/(b-a)

CDF F(x) = (x-a)/(b-a)

Sums

1) If a continuous random variable x follows rectangular distribution in the range (2,7), find the probabilities: i) P(2.5<=x<=4) ii) P(x<=3.4) iii) P(x>6) iv) P(x=4.5)

A: i) P(2.5<=x<=4) = (4-2.5)/(7-2) = 0.3
(ii) P(x<=3.4) = F(3.4) = (3.4-2)/(7-2) = 0.28
(iii) P(x>6) = 1-P(X<=6) = 1 - (6-2)/(7-2) = 1 - 0.8 = 0.2
(iv) P(x=4.5) = 0. (No probability for x at a single point

Normal approximation to binomial

If p is small and n is large

=> Bin(n,p) -> N(µ, σ)
=> µ = np, σ = √(npq)

z = (x-µ)/σ ~ N(0,1)

1) A fair coin is tossed 400 times. Using normal approximation, find the probability of obtaining (i) exactly 200 heads (ii) less than 210 heads (iii) between 190 and 210 heads, both inclusive. Given that the area under the standard normal curve between z=0 and z=.05 is .0199, between z=0 and z=.95 is .3289, between z=0 and z=1.05 is .3531.

A: n=400, p=1/2 => µ = np = 200, σ = √(npq) = 10

P(X=200)
=> P(199.5<=Y<=200.5)
=> P(-0.05<=z<=0.05) = .0199 * 2 = 0.0398 [i]

P(X<210)
=> P(Y<=209.5)
=> P(z<=0.95) = 0.5+0.3289 = 0.8289 [ii]

P(190<=X<=210)
=> P(189.5<=Y<=210.5)N
=> P(-1.05<=z<=1.05) = 0.3531 * 2 = 0.7062 [iii]

Normal approximation to Poisson

Poisson (λ) -> Nor(µ, σ)

=> µ = λ, σ = √λ

=> Y~Nor(λ, √λ)

z = (y-µ)/σ

USE ONLY IF λ>10

2) A variable x follows Poisson distribution with mean 16. Find the probability P(x>=20). Given F(0.875) = .8092 where F(x) is the standard normal distribution function.

A: µ = 16, σ = 4

P(X>=20)
=> P(Y>=19.5)
=> P(z>=0.875)
=> 1 - P(z<0.875)
=> 1 - 0.8092 = 0.1908

Exponential Distribution

f(x) = λe^(-λx) when x>=0, 0 otherwise. (PDF)

F(x) = P(X<=x) = ∫λe^(-λx)dx = 1-e^(-λx) (CDF)

Mean = E(x) = 1/λ
Variance = E(x^2) - E^2(x) = 1/λ^2
Parameter = λ

Memoryless property: P(X1|X2) = P(X1)

1) The lifetime of a light bulb follows exponential distribution with a mean of 1000 hours. If the bulb is guaranteed to last at least 900 hours, find the probability that it will satisfy the guarantee.

A: Mean = 1000 => λ = 1/1000

P(X>=900) => 1 - P(X<900) = 1 - F(900) = 1 - (1-e^(-λx) = e^(-λx) = e^(-0.9) = 0.4066