Compound Events: Probability of Complement of An Event

Probability part III: Compound Events
Compound events can be formed by combining two or more events. The combinig can be done by union or intersection. For example in the experiment of tossing two dice, suppose A is the event of the appears of dice eyes are the prime numbers, and B is the event of the appears of dice eye are the odd numbers, then event of the appear of dice eyes are prime umbers or odd umbers are the compound events that can be written by A∪B, so does event of the appear of dice eyes are prime numbers and odd numbers are the compound events that can be written by A∩B.

Probability of complement of an event
An experiment may consist of several events. If we establish a spesific event suppose we callit A, then the events outside A which are the results of these experiments, known as the complement of event A. Because of one event is the set (which is the set of one or more sample points) then, the notation for the complement of event is equal to the notation of the complement of the set. So the complement of event A is denoted by Ᾱ. The union of event A and the complement of event A is equal to the sample space.

chance of complement event

In a dice throwing experiment, the sample space S={1,2,3,4,5,6}. Suppose A is the event of the appears of number are less than 3, then A ={1,2} and Ᾱ={3,4,5,6}, then P(A)=2/6;P(Ᾱ)=4/6.
Note that P(A)+P(Ᾱ)=1 or P(Ᾱ)=1-P(A).

Suppose K is an event in an experiment.
The probability of complement of event K is P(K ̅)=1-P(K)

Example 1: Suppose a pairs of dice are thrown at same time. X states the event of the sum of dice eyes that appear are prime number.
A. Determine the probability of X.
B. Determine the probability of the sum of dice eyes that appear aren’t prime number. The amount of members of the sample space =62=36.
The explanation:
a. Suppose (x,y) states couple of dice eye I and II that appear. X is the event of the sum of dice eye that appear are prime numbers then:

X={((1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(5,6),(6,5))}

Because the amount members of the event is n(X)=15, then P(X)=15/36.
b. The probability of events of sum of dice eye that appear are not prime numbers is:

P(X ̅)=1-P(X)=1-15/36=21/36
Exercise Competency Test 1
Instruction: Choose the one correct answer of each question.
1. Suppose probability of somebody shoots right on target is equal to 0.6. Thus the probability of that somebody shoots on the outside of the target is equal to …
A. 0.3 B. 0.4 C. 0.5 D. 0.7 E. 0.8
correct: B, the explanation:

P(Ᾱ)=1-P(A)
P(Ᾱ)=1-0.6=0.4
2. In throwing of 3 coins once. If the coins consist of image side and numeral side, then the probability of the emerging of at least one image side is equal to …
A. ⅛ B. ⅜ C. ⅝ D. ¾ E. ⅞
correct: E, the explanation:
We don’t need to count each probability. Probabilities of one image, two and three ones don’t need to be counted!
We need to count the probability of complement of that event first.
The sample space of throwing a coin once is equal to 2.
The sample space of throwing k coins once is equal to 2k.
Thus the sample space of throwing 3 coins once is equal to n{S}=23=8.
Suppose E states image side and N states numeral side. Suppose A consist of appearances at least an image side and Ᾱ is complement of event A. Thus Ᾱ consist of appearance zero image sides. Thus

Ᾱ={NNN}→n{Ᾱ}=1
P(Ᾱ)=⅛

The probability of event A is equal to

P(A)=1-P(Ᾱ)
P(A)=1-⅛=⅞
3. A marbel is withdrawn randomly from a bag. The bag contains 6 red marbles, 7 white marbles and 2 blue marbles. Probability of the withdrawn marbles aren’t red marbles is equal to …
A. ⅕ B. ⅖ C. ⅗ D. ⅘ E. 5/7
correct: C, the explanation:
White and blue aren’t red. Clear!

chance of white blue marble

4. In a box there are 6 marbles red and 5 white ones. From the box three marbles are taken once. The probability of at least one red marble of three marbles taken is equal to …
A. 32/33 B. 31/33 C. 10/11 D. 3/33 E. 1/33
correct: B, the explanation:
This question is enable to be solved such as the explanation of question number two.
Suppose W denote a red marble and W denote a white one.

Ᾱ={WWW}

Remember the basic combination formula

chance of no red marble taken

The probability of at least one red marble taken is equal to …

chance of at least certain number of marbles taken

RELATED POSTs

Leave a Reply

Your email address will not be published.