Lecture Notes MTH302 Before MTT Myers.docx

Definition:
Example: 2.2: An experiment consists of flipping a
coin and then flipping it a second time if a head occurs.
If a tail occurs on the first flip, then a die is tossed
once. Write the sample space of the experiment.
S = {HH, HT, T1, T2, T3, T4, T5, T6}

Que 2.5: An experiment consists of tossing a die
and then flipping a coin once if the number on the die
is even. If the number on the die is odd, the coin is
flipped twice. Using the notation 4H, for Example:, to
denote the outcome that the die comes up 4 and then
the coin comes up heads, and 3HT to denote the
outcome that the die comes up 3 followed by a head
and then a tail on the coin, construct the sample space
S.

Que 2.5: An experiment consists of tossing a die and
then flipping a coin once if the number on the die is even.
If the number on the die is odd, the coin is flipped twice.
(a) What is the probability of the event that the
number on the die is less than 3?
(b) What is the probability of the event that the two
tails occur?
Example::

Que. 2.7 Four students are selected at random from a
chemistry class and classified as male or female. List the
elements of the sample space S1, using the letter M for male
and F for female. Define a second sample space S2 where the
elements represent the number of females selected.

Definition 2.7: A permutation is an arrangement of all or
part of a set of objects.
Theorem 2.1: The number of permutations of n objects is n!.
Example::

Theorem 2.3: The number of permutations of n objects
arranged in a circle is (n − 1)!.

Example: 2.21: In how many ways can 7 graduate students
be assigned to 1 triple and 2 double hotel rooms during a
conference?
Que. A college plays 12 football games during a season. How
many ways can the team end the season with 7 wins, 3 loses,
and 2 ties?
Example: A young boy asks his mother to get 5 Game-Boy
cartridges from his collection of 10 arcade and 5 sports games.

How many ways are there that his mother can get 3 arcade and
2 sports games?
Example: How many different letter arrangements can be
made from the letters in the word STATISTICS?
Example: In a random arrangement of the letters of the
word ‘COMMERCE’, find the probability that all the vowels
come together.
Example: ‘7’ persons are seated on ‘7’ chairs around a table.
The probability that three specified persons are always sitting
next to each other is:
(a)1/4 (b)1/5 (c)1/6 (d) 1/3.

Example: In a poker hand consisting of 5 cards, find the
probability of holding 2 aces and 3 jacks.
Q. A man is dealt/ given 5 hearts cards from a pack of 52 cards. If he is
given 4 more additional cards, then the probability that at least one of
the additional cards is also a heart is:
(a) 39C4/52C5 (b)1-(39C4/47C4) (c)39C4/47C4 (d) None of
these.

Q. In how many ways can 4 boys and 5 girls sit in a row if the boys and
girls must alternate?
Example: An MBA applies for a job in two firms X and
Y. The probability of his being selected in firm X is 0.7,
and being rejected at firm Y is 0.5.The probability of at
least one of his applications being rejected is 0.6. What
is the probability that he will be selected in one of the
firms?

(a) 0.2 (b) 0.8 (c) 0.7 (d)
None of these.
Example: A,B,C are three mutually exclusive and
exhaustive events associated with the random
experiment. Given that 𝑃(𝐵) =
3
2
𝑃(𝐴) and 𝑃(𝐶) =
1
2
𝑃(𝐵) then 𝑃(𝐴) =?
(a) 4/9 (b) 9/13 (c) 4/13 (d)
None of these.
Que. Suppose that in a senior college class of 500 students it
is found that 210 smoke, 258 drink alcoholic beverages, 216
eat between meals, 122 smoke and drink alcoholic
beverages, 83 eat between meals and drink alcoholic
beverages, 97 smoke and eat between meals, and 52 engage
in all three of these bad health practices. If a member of this
senior class is selected at random, find the probability that
the student
(a) Smokes but does not drink alcoholic beverages;

(b) Eats between meals and drinks alcoholic beverages but
does not smoke;
(c) Neither smokes nor eats between meals.
Que. In a high school graduating class of 100 students, 54 studied
mathematics, 69 studied history, and 35 studied both mathematics and
history. If one of these students is selected at random, find the
probability that (a) the student took mathematics or history; (b) the
student did not take either of these subjects; (c) the student took
history but not mathematics.
Que. If five dice are rolled, what is the probability of getting four of a
kind?
𝐶(5,4) ∗ 6 ∗ 5
65

Example: The probability that a regularly scheduled
flight departs on time is P(D)=0.83; the probability that
it arrives on time is P(A)=0.82; and the probability that
it departs and arrives on time is
P(D ∩ A)=0.78. Find the probability that a plane
(a) arrives on time, given that it departed on time,
(b) departed on time, given that it has arrived on time.
(c) arrives on time, given that it has not departed on
time

Q. Which of the following statements is/are correct?
(i) If events 𝐴 and 𝐵 are mutually exclusive then
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)
(ii) If events 𝐴 and 𝐵 are mutually exclusive then 𝑃(𝐴 ∪
𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
(iii) If events 𝐴 and 𝐵 are mutually independent
then 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵).
(a) option (i) is correct only (b) option (ii) is
correct only
(c) both options (ii)and (iii) are correct only
(d) option (iii) is correct only.
Example: Suppose that we have a fuse box containing 20
fuses, of which 5 are defective. If 2 fuses are selected at
random and removed from the box in succession without
replacing the first, what is the probability that both fuses are
defective?

Example: One bag contains 4 white balls and 3 black
balls, and a second bag contains 3 white balls and 5
black balls. One ball is drawn from the first bag and
placed unseen in the second bag. What is the
probability that a ball now drawn from the second bag
is black?

Example: An electrical system consists of four components
as illustrated in Figure. The system works if components A
and B work and either of the components C or D works. The
reliability (probability of working) of each component is also
shown in Figure. Find the probability that
(a) the entire system works and
(b) the component C does not work, given that the
entire system works. Assume that the four components
work independently.

Example:
Theorem of total probability or the rule of
elimination

Example:
If a product was chosen randomly and found to be
defective, what is the probability
that it was made by machine B3?

The random variable for which 0 and 1 are chosen to
describe the two possible
values is called a Bernoulli random variable.
Example:
Example:
A random variable is called a discrete random
variable if its set of possible outcomes is countable.

When a random variable can take on values on a
continuous scale, it is called a continuous random
variable.
Probability Distribution Function
Example:

Note: When X is a continuous random variable then

𝑃(𝑎 < 𝑋 < 𝑏) = 𝑃(𝑎 ≤ 𝑋 < 𝑏) = 𝑃(𝑎 < 𝑋 ≤ 𝑏) = 𝑃(𝑎
≤ 𝑋 ≤ 𝑏)
That is, it does not matter whether we include an
endpoint of the interval or not. This is not true, though,
when X is discrete.
In fact, if ‘X’ is a continuous random variable then
𝑃(𝑋 = 𝑐) = 0. where 𝑐 is any constant.
As an immediate consequence of the above
Definition, one can write the two results
if the derivative exists.
Example:

Also find the cumulative distribution function of the
random variable X.
Using F(x), verify that f(2) = 3/8.
Example:

Also find the cumulative distribution function of the
random variable M.
Example:

(c) Find F(x), and use it to evaluate P(0 < X ≤ 1).
Example:

Example: If 𝑝(𝑥) = {
𝑥
15
; 𝑖𝑓 𝑥 = 1,2,3,4,5
0 ; 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
then find (i)
P(X=1 or 2) (ii) 𝑃 (
1
2
< 𝑋 <
5
2
|𝑋 > 1)
(a) 1/7 (b) 2/15 (c) 1/5 (d)
None of these.

Find the conditional distribution of X, given that Y = 1, and
use it to determine P(X = 0 | Y = 1).

Remarks:
𝑬[𝒂𝟏𝒇𝟏(𝑿) + 𝒂𝟐𝒇𝟐(𝑿) + ⋯ … + 𝒂𝒏𝒇𝒏(𝑿)] =
𝒂𝟏𝑬[𝒇𝟏(𝑿)] + 𝒂𝟐𝑬[𝒇𝟐(𝑿)] + ⋯ … + 𝒂𝒏𝑬[𝒇𝒏(𝑿)]
𝑬(𝒄) = 𝒄 where ‘𝒄’ is any constant.
Example:

𝐶𝑜𝑣. (𝑋, 𝑌) = 𝜎𝑋𝑌 = 𝐸[(𝑋 − 𝜇𝑋)(𝑌 − 𝜇𝑌)] = 𝐸[𝑋𝑌 −
𝜇𝑌𝑋 − 𝜇𝑋𝑌 + 𝜇𝑋𝜇𝑌 ]
= 𝐸[𝑋𝑌] − 𝐸[𝜇𝑌𝑋] − 𝐸[𝜇𝑋𝑌] + 𝐸[𝜇𝑋𝜇𝑌] = 𝐸[𝑋𝑌] −
𝜇𝑌𝐸[𝑋] − 𝜇𝑋𝐸[𝑌] + 𝜇𝑋𝜇𝑌
= 𝐸[𝑋𝑌] − 𝜇𝑌𝜇𝑋 − 𝜇𝑋𝜇𝑌 + 𝜇𝑋𝜇𝑌 = 𝐸[𝑋𝑌] − 𝜇𝑌𝜇𝑋 =
𝐸[𝑋𝑌] − 𝐸[𝑋]𝐸[𝑌]
Note: If the random variables 𝑿 and 𝒀 are
independent then 𝐶𝑜𝑣. (𝑋, 𝑌) = 𝜎𝑋𝑌 = 0

Since, if the random variables 𝑿 and 𝒀 are
independent then 𝐸[𝑋𝑌] = 𝐸[𝑋]𝐸[𝑌].
Example:
Find the covariance of ‘X’ and ‘Y’.

Que. If 𝑋1 and 𝑋2 are two random variables, 𝑎1 and 𝑎2 are two
constants then 𝑉(𝑎1𝑋1 + 𝑎2𝑋2) =?
(a) 𝑎1
2
𝑉(𝑋1) + 𝑎2
2
𝑉(𝑋2) − 2𝑎1
2
𝑎2
2
𝐶𝑜𝑣(𝑋1, 𝑋2) (b) 𝑎1
2
𝑉(𝑋1) +
𝑎2
2
𝑉(𝑋2) + 2𝑎1
2
𝑎2
2
𝐶𝑜𝑣(𝑋1, 𝑋2)
(c) 𝑎1
2
𝑉(𝑋1) + 𝑎2
2
𝑉(𝑋2) − 2𝑎1𝑎2𝐶𝑜𝑣(𝑋1, 𝑋2) (d) 𝑎1
2
𝑉(𝑋1) +
𝑎2
2
𝑉(𝑋2) + 2𝑎1𝑎2𝐶𝑜𝑣(𝑋1, 𝑋2).
Remark: If 𝑋1, 𝑋2, 𝑋3, … … . . , 𝑋𝑛 are ‘𝑛’ random variables and
𝑎1, 𝑎2, 𝑎3, … … . . , 𝑎𝑛 are ‘𝑛’ constants then

𝑉(𝑎1𝑋1 + 𝑎2𝑋2 + ⋯ … . +𝑎𝑛𝑋𝑛) = 𝑉(∑𝑎𝑖𝑋𝑖)
= ∑𝑎𝑖
2
𝑋𝑖
2
+ 2∑∑𝑎𝑖𝑎𝑗𝐶𝑜𝑣(𝑋𝑖, 𝑋𝑗)
BINOMIAL DISTRIBUTION
EXAMPLE:
Where Does the Name Binomial Come From?

Example: In a binomial distribution consisting of five
independent trials, probabilities of one success and
three failures are 0.4096 and 0.2048 respectively. Then
what is the probability of success and failure in a single
trial?
(a) 0.5 and 0.5 resp. (b) 0.2 and 0.8 resp. (c)
0.4 and 0.6 resp. (d) 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒.

Negative Binomial Distribution
If there are ‘𝒙’ number of trials for the ‘𝒌’ number of
successes then obviously in the first
(𝒙 − 𝟏)number of trials there will be (𝒌 − 𝟏)number
of successes (we denote this event by 𝑬𝟏)
and in the 𝒙𝒕𝒉
trial there will be the last 𝒌𝒕𝒉
success
(we denote this event by 𝑬𝟐).
Now, 𝑷(𝑬𝟏) = (𝒙−𝟏
𝒌−𝟏
)𝒑𝒌−𝟏
𝒙𝒙−𝟏−(𝒌−𝟏)
= (𝒙−𝟏
𝒌−𝟏
)𝒑𝒌−𝟏
𝒙𝒙−𝒌
and 𝑷(𝑬𝟐) = 𝒑. Therefore the required probability
is given by 𝑷(𝑬𝟏 ∩ 𝑬𝟐) = 𝑷(𝑬𝟏)𝑷(𝑬𝟐) =
(𝒙−𝟏
𝒌−𝟏
)𝒑𝒌−𝟏
𝒙𝒙−𝒌
∗ 𝒑 = (𝒙−𝟏
𝒌−𝟏
)𝒑𝒌
𝒙𝒙−𝒌
.
EXAMPLE:

Result:
Proof: We have 𝒃(𝒙, 𝒏, 𝒑) = (𝒏
𝒙
)𝒑𝒙
𝒒𝒏−𝒙
=
𝒏!
𝒙!(𝒏−𝒙)!
𝒑𝒙
𝒒𝒏−𝒙
[use 𝒏𝒑 = 𝝁 or 𝒑 =
𝝁
𝒏
]
=
𝒏(𝒏 − 𝟏)(𝒏 − 𝟐)(𝒏 − 𝟑) … … … … (𝒏 − 𝒙 + 𝟏)(𝒏 − 𝒙)!
𝒙! (𝒏 − 𝒙)!
(
𝝁
𝒏
)
𝒙
(𝟏
−
𝝁
𝒏
)
𝒏−𝒙
=
𝒏(𝒏 − 𝟏)(𝒏 − 𝟐)(𝒏 − 𝟑) … … … … (𝒏 − 𝒙 + 𝟏)
𝒙!
(
𝝁
𝒏
)
𝒙
(𝟏
−
𝝁
𝒏
)
𝒏−𝒙

=
𝒏
𝒏
(𝒏 − 𝟏)
𝒏
(𝒏 − 𝟐)
𝒏
(𝒏 − 𝟑)
𝒏
… … … …
(𝒏 − 𝒙 + 𝟏)
𝒏
𝒙!
(𝝁)𝒙
(𝟏 −
𝝁
𝒏
)
𝒏−𝒙
=
𝟏 (𝟏 −
𝟏
𝒏
) (𝟏 −
𝟐
𝒏
) (𝟏 −
𝟑
𝒏
) … … … … (𝟏 −
𝒙 − 𝟏
𝒏
)
𝒙!
(𝝁)𝒙
(𝟏 −
𝝁
𝒏
)
𝒏
(𝟏 −
𝝁
𝒏
)
𝒙
Now taking limit as 𝒏 → ∞ on both side and using the
result that 𝐥𝐢𝐦
𝒏→∞
(𝟏 +
𝒌
𝒏
)
𝒏
= 𝒆𝒌
we get
𝒃(𝒙, 𝒏, 𝒑) →
𝝁𝒙𝒆−𝝁
𝒙!
= 𝒑(𝒙, 𝝁). Hence Proved.

Proof:
Areas under the normal curve

Applications of Normal Distribution:

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