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Probablity is the study of randomness and uncertainty. Take an example of throwing an unbiased coin.
If we knew more about the:
we would be able to definitively say whether the coin would come up heads or tails. However, in the absence of all that information, we cannot predict the outcome of the coin flip.
When we say that something is random, we are saying that our knowledge about the outcome is limited, so we can’t be certain what will happen.
The probability of an outcome is interpreted as the long-run proportion of the time that the outcome would occur, if the experiment were repeated indefinitely. That is, probability is long-term relative frequency.
Ex. Construct a sample space for the experiment that consists of rolling a single die. Find the events that correspond to the phrases “an even number is rolled” and “a number greater than two is rolled.”
The outcomes could be labeled according to the number of dots on the top face of the die. Then the sample space is the set S={1,2,3,4,5,6}.
The outcomes that are even are 2, 4, and 6, so the event that corresponds to the phrase “an even number is rolled” is the set {2,4,6}, which it is natural to denote by the letter E. We write E={2,4,6}.
Similarly the event that corresponds to the phrase “a number greater than two is rolled” is the set T={3,4,5,6}, which we have denoted T.
VENN DIAGRAMS
Let’s solve a problem using logic and not the formulas:
Ex. In a class of 50 students, 30 take philosophy, 23 take sociology and 12 take both. What is the probablity that a randomly selected student-
Making a venn diagram for the given data:
30 take philosophy and 12 take both, so the ones who only take philosophy will be 30-12 i.e. 18, as represented in the venn diagram. Similarly, the ones who take only sociology will be 11.
The total students who take subjects are -18+12+11= 41, which means 9 are left.
When taking union of 2 sets, adding them directly will cause an error. By adding the sets directly we include the intersection twice, therefore we must subtract the intersection once.
The addition rule states:
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
It is illustrated in the venn diagram below:
Ex. A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a king or a club?
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
𝑃(𝑘𝑖𝑛𝑔 𝑜𝑟 𝑐𝑙𝑢𝑏) = 𝑃(𝑘𝑖𝑛𝑔) + 𝑃(𝑐𝑙𝑢𝑏) − 𝑃(𝑘𝑖𝑛𝑔 𝑜𝑓 𝑐𝑙𝑢𝑏𝑠) = ((4/52) + (13/52) − (1/52)) = 16/52 = 4/13
If you flip a coin twice, you can model the possible outcomes using a tree diagram or an outcome table.
Outcome table:
Flip1 | Flip 2 | Event |
H | H | HH |
H | T | HT |
T | H | TH |
T | T | TT |
We can also show this using a 2-D grid:
Mutually exclusive (or disjoint) events are events whose outcomes cannot occur at the same time.
Ex. (a) Turning left and turning right are Mutually Exclusive (you can’t do both at the same time)
(b) Tossing a coin: Heads and Tails are Mutually Exclusive
(c) Cards: Kings and Aces are Mutually Exclusive
The events which cannot be mutually exclusive:
Ex. (a) Turning left and scratching your head can happen at the same time
(b) Kings and Hearts, because we can have a King of Hearts!
When two events (call them “A” and “B”) are Mutually Exclusive it is impossible for them to happen together, which implies the intersection of these events is zero.𝑃(𝐴 ∩ 𝐵) = 0 From addition rule –
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
For mutually exclusive events-
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
Ex. If the probability of:
scoring no goals (Event “A”) is 20%
scoring exactly 1 goal (Event “B”) is 15%
Find the probablity of scoring no goals or 1 goal
The probability of scoring no goals and 1 goal is 0 (Impossible)
The probability of scoring no goals or 1 goal is 20% + 15% = 35%
Can be written as 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) (mutually exclusive events) 𝑃(𝐴 ∪ 𝐵) = 20% + 15% = 35% The probablity of scoring no goals or 1 goal is 35%.
When the likelihood of happening of two events are same they are known as equally likely events. Example-
Ex. In the experiment of throwing a die, where
G: the event of getting a prime number
H: the event of getting an even number
Probability of 𝐺 = 1/2
Probability of H = 1/2
The two events G and H are equally likely.
But they are not mutually exclusive since both events occur when 2 apperas on the die.
Some problems may require the need to geometry (areas in plane) to solve.
Ex Find the probability that a randomly chosen point in the figure lies in the shaded region.
Total area = 𝜋(5)^{2} = 25𝜋
Area of shaded region = 𝜋(5)^{2} − 𝜋(5 − 3)^{2} = 21𝜋
Required probability = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛/𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 = 21𝜋/25𝜋 = 0.84
Ex. A store has six different fitness magazines and three different news magazines. If a customer buys three magazines at random, find the probability that the customer will pick two fitness magazines and one news magazine.
There are 26C or 15 ways to select two fitness magazines from six fitness magazines.
26C 6!/(6−2)!2! = 6!/4!2! = 15
There are 13C or three ways to select one magazine from three news magazines.
13𝐶 = 3!/(3−1)!1! = 3!/2!1! = 3
Using fundamental counting principle, there are 15×3 or 45 ways to select two fitness magazines and one news magazine.
Ex. Probability of 3 Heads in a Row For each toss of a coin a Head has a probability of 0.5 And so the chance of getting 3 Heads in a row is 0.125.
Ex. Your boss (to be fair) randomly assigns everyone an extra 2 hours work on weekend evenings between 4 and midnight.
What are the chances you get Saturday between 4 and 6?
Day: there are two days on the weekend, so P(Saturday) = 0.5
Time: you want the 2 hours of “4 to 6”, out of the 8 hours of 4 to midnight: P(“4 to 6”) 2/8 0.25
P(Saturday and “4 to 6”) P(Saturday) P(“4 to 6”)
=0.5×0.25
=0.125
Or a 12.5% Chance.
Ex. We have 2 blue marbles and 3 red marbles. What are the chances of drawing out 2 blue marbles.
Let’s draw a tree diagram we learnt in previous section.
There is a 2/5 chance of pulling out a Blue marble, and a 3/5 chance for Red:
We can go one step further and see what happens when we pick a second marble:
It is a 2/5 chance followed by a 1/4 chance of drawing out a blue marble ((2/5) x (1/4)) = 1/10
Notation:
In our marbles example Event A is “get a Blue Marble first” with a probability of 2/5:
P(A) = 2/5
And Event B is “get a Blue Marble second” … but for that we have 2 choices:
P(B|A) (conditional probability) means “Event B given Event A”
In other words, event A has already happened, now what is the chance of event B?
And in our case: P(A) = 1/4 So the probability of getting 2 blue marbles is:
And we write it as:
𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴) × 𝑃(𝐵|𝐴)
𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐵) × 𝑃(𝐴|𝐵)
or we can write it as:
𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨) × 𝑷(𝑩|𝑨)
𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑩) × 𝑷(𝑨|𝑩)
We know that for independent events 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 𝑃(𝐵) Therefore,
𝑃(𝐵|𝐴) = 𝑃(𝐵)
𝑃(𝐴|𝐵′ ) = 𝑃(𝐴)
𝑃(𝐵|𝐴′ ) = 𝑃(𝐵)
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