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Complimentary Events

In online statistics help , Probability is way of expressing object or the individual event. Probability theory that is used extensively in such areas of study as mathematics, statistics, and finance etc.Therotical probability is the likeliness of an event happening based on all the possible outcomes.Experimental probability of an event is the ratio of the number of times the event occurs to the total number of trials.

For a given sample (s) the probability of obtaining A is called p (A). The probability of not obtaining A is called p’ (A) this is called complimentary event of A of sample s.The event A and not (A) are called mutually exclusive or exhaustive events. This case is used in probability in solving some critical or lengthy problem; this will be short method and can be completed easily.to solve easily we can use online probability calculator

Complimentary event can be denoted as p’ (A) which is 1--p (A).

Complements of intersections and unions are made by demorgan law:

(A U B) ‘ = A’ n B’

The not happening events of A or B in a sample is equal to not happening events of A and B.

(A n B)’ =A’ U B’

The not happening events of A and B in a sample is equal to not happening events of A or B.

Example: The probability of getting blue ball from a bag is 1/5 and white ball is 1/4 find the complimentary event of it?

Given: Probability of getting blue ball from a bag is p (B) =1/5

Probability of getting white ball from a bag is p (W)=1/4

Probability of not getting blue ball is p’ (B)

p’ (B)=1-p (B)

=1-1/5

= (5-1)/5

Probability of not getting blue ball is p’ (B) is 4/5

Probability of not getting blue ball is p’ (w)

p’ (W)= 1-p(W)

=1=1/4

=(4-1)/4

Probability of not getting blue ball is p’ (w) is 3/4

Note: The probability of an event should be less than 1 and

A and B are called complementary events. This may be denoted as:

P (A’) = P (B) (recall in sets that A’ is the complement of A)

P (A) = P (B’)

We can generally state that: P (A) + P(A ’ ) = 1

We can verify the answer by checking the result probability.

Bayes’s Theorem

It relates through conditional probability and it’s used in many areas. It involves in both the events like probability of occurring and non occurring events to give the probability of occurring of A when B has occurred.

Let A1,A2,……..Ai be a collection of mutually exclusive and exhaustive events with p(Ai)>0 for all i.

Then for any other event B with p (B)>0 we have:

Complimentary Events

Solved Problems

1) Bag contains blue and green cards .The probability of drawing blue card is 1/5 what is the probability of drawing green cards?

Solution: Event of drawing blue card is B

Event of drawing green card is G

The probability of drawing blue card =P (B)

The probability of drawing Green card =P(G)

P(B) is the probability of drawing a blue card which is also the same as the probability of not drawing a Green card (Since the cards are either green or blue)

P (G) = P (of not getting blue card)

=P’ (B)

= (1-P (B))

=1-1/5

= (5-1)/5

=4/5

Verification: A and B are called complementary events. This may be denoted as:

P (A’) = P (B) (recall in sets that A’ is the complement of A)

P (A) = P (B’)

We can generally state that: P (A) + P(A ’ ) = 1

1/5+4/5=5/5=1

Hence proved

2) The drug analysis of a factory, gives a positive result 97% of the time when the drug is present:

Gives a positive result 0.4% of the time if the drug is not present (“false positive”). What is the probability that the drug is not present and you have a positive test?

Solution: P (positive test drug present) =0.97

P (positive test drug not present) =0.004

Let’s assume that the drug is present in 0.5% of the population (1 out of 200 people).

P (drug present)=0.005

P (drug not present)=1-P(drug present)=0.995

P(drug is not present|positive test)=????

Bayes’s Theorem gives:

P(drug is not present|positive test)=

p(test positive | drug not present) (drug not present)
p(test positive | drug not present) (drug not present) + p(test positive | drug present) (drug present)
(drug not present | test positive) = (0.004)(1-0.005)
(0.004)(1 0.005)+ (0.97)(0.005)

(drug not present | test positive) =0.45

Thus there is a 45% chance that the test comes back positive even if you are drug free!

The real life consequence of this large probability is that drug tests are often administered twice.

Students can avail the help with Probability homework problems online.