"Mathematics is a easy subject but maximum find them so hard to learn and remember Mathematics functions and rules. That all because we not give time to understand basics of Mathematics rule. So learn here the basic rules step by step."
PROBABILITY
"Probability is a numerical description of possibilities that an event can be occur."
The probability of an event is a number between 0 and 1. '0' indicates impossibility of the event and '1' indicates certainty to find the probability of a single event to occur we need to know the total number of possible outcomes.
Probability can be written as decimal value or in percentage.
The possibility of an event will be written as P ( E ) .
1. If P ( A ) › P (B) ; then event A has higher chance of occurring then event B.
If P ( A ) = P (B) ; then event A and B are equally likely to occur.
2. If an event may happen 'X' ways and fails to happen in 'Y' ways in a trial ,
Then the probability to event happening P ( E ) = X∕(X+Y) , and
The probability to event fails to happening = 1- P (E)
= 1- ⟨ X∕(X+Y) ⟩
FORMULAS :
1. Probability of an event occur P ( E ) = ( Number of Favorable Outcomes )/ ( Total Number of Outcomes)
2. Probability that the event will not occur P ( E' ) = 1- P ( E )
3. P ( E ) + P ( E' ) = 1
SOME EXAMPLES FOR UNDERSTANDING :
Example 1.
If we tossing a coin then :-
(i). Probability to find Head P (H) = 1/2 = 0.5 = 50 %
(!!). Probability to find Tail P (T) = 1/2 = 0.5 = 50 %
Here, P (H) = P (T)
Example 2 .
If coin tossing 2 times then,
Total possible events are HH, TT, HT, TH = 4 events
Hence,
(i). Possibilities to find 2 Head P (H2) = 1/4
(ii). Possibilities to find 2 Tail P (T2) = 1/4
(iii). Possibilities to find one Head and one Tail P (HT) = 2/4 = 1/2
Example 3 .
A dice rolled one time then total possible outcomes will be 1,2,3,4,5,6 = 6
Hence,
(i). Possibilities to find 1 = P (1) = 1/6
(ii). Possibilities to find 1or 5 = P (1or 5) = 2/6 = 1/3
(iii). Possibilities to find 1or 3 or 5 = P ( 1or 3 or 5 ) = P ( ODD )= 3/6 = 1/2
(iv). Possibilities to find 2or 4 or 6 = P ( 2or 4 or 6 ) = P ( EVEN )= 3/6 = 1/2
Example 4
There are 6 books in a bag. 2 are English, 3 are Maths, 1 is science. What is possibility to get an English book.
Solution :
Here , No. of favorable chances to get English book = 2
Total number of outcomes = 2+3+1 = 6
Hence,
P ( English ) = 2/6 = 1/3
Example 5 .
Two coins are tossed 300 times and find -
2 Heads = 100 Times,
1Head = 50 Times,
No Head = 150 Times.
Then;
Probability to get 2 Heads = P (E1) = 100/300 = 1/3
Probability to get 1 Head = P (E2) = 50/300 = 1/6
Probability to get No Head = P (E3) = 150/300 = 1/2
Example 6 .
The record of a weather station shows that out of past one year , its weather forecast were correct 225 times. Then--
(!). The probability that on a given day it will be correct P (E)
= (No. of days when the forecast was correct) / (Total no. of days of forecast)
= 225 / 365 = 0.62 = 62%
(ii). The probability when the forecast will not br correct P (E')
= 1 - P (E)
= 1-0.62 = 0.38 = 38%
OR
= (365 - 225) days / 365 days = 140 / 365 = 0.38 = 38%
NOTE : Always P (E) + P (E') = 1 or 100%
LAWS OF PROBABILITY
There are three important Laws of Probability, they are under below :--
1. Addition Law of Probability
2. Multiplication Law of Probability
3. Law of Conditional Probability
1. Addition Law of Probability :
" The probability of any event is the sum of the probability of its mutually exclusive events (sub events). "
If the two events "A" and "B" are mutually exclusive and,
P (A) = Probability of occurrence of event A ,
P (B) = Probability of occurrence of event B.
Then the probability of occurrence of the compound (Total or whole) event A+B is -
P ( A+B ) = P (A)+P (B)
That we can also explain following way :-
P ( A∪B ) = P(A)+ P(B)-P ( A∩B )
Here A and B are disjoint , Hence P ( A∩B ) = 0
So that,
P ( A∪B ) = P(A)+ P(B) (For exclusive events)
EXAMPLE :
In a bag contains 200 balls, in which 80 red balls, 50 white balls, 70 blue balls.
If we draw one ball randomly from bag. Then-
(i). The probability that will be red (PR ) = 80/200 = 0.4
(ii). The probability that will be white (PW ) = 50/200 = 0.25
(iii). The probability that will be blue (PB ) = 70/200 = 0.35
(iv). So, that the probability that it will be red or white or blue
= PR + PW + PB = 0.4 + 0.25 + 0.35 = 1
2. Multiplication Law of Probability :
" If a compound event be made up of number of separate and independent events, and occurrence of the compound event be the result of each of these sub-events happening, the probability of occurrence of the compound event is the product of the probabilities that each of these sub events will happen."
So, where it is possible for two or more sub events to occur together, the probability they will do so, is the product of their individual probability.
P ( A . B ) = P(A) + P(B)
3. Law of Conditional Probability :
When the events are not independent but the nature of the dependence is known, the law of conditional probability may be use.
This Law states that " the probability that both of two dependent events will occur is the probability of the first multiply by the probability that if the first has occurred the second will also happen."
If two dependent events are A and B then the probability that both of two dependent events A and B will occur is given as below-
P (B) x P (A/B)
Where P (B) = Probability of occurring of event B ,
P (A/B) = Probability that A happens given that B has happened.
( Here probability of A conditioned by B )
TO SOLVE ABOVE LAW FOLLOWING MUST BE LEARN :-
( A ). Factorial Notation :
1. n! = The product of first "n" natural numbers.
n! = n x (n-1) x (n-2) x ... x 1
n! = n x (n-1)!
0! = 1
EXAMPLE :
1. 4! = 4 X 3 X 2 X 1 = 24
2. 5! = 5 X 4 X 3 X 2 X 1 = 120
3. 6! = 6 X 5 X 4 X 3 X 2 X 1 = 720 etc.
4. 6! = 6 x 5 x 4! or 6 x 5! etc.
2. If in a problem we select 'r' objects out of 'n' distinguishable objects and arranging them in order, then we can notation as below :-
nPr = n (n-1) (n-2) ... (n-r+1) = n!/ (n-r)!
( B ). Combinations :
" A combination is the number of ways in which a certain number of articles are selected from a lot, without regard to order in which they are drawn."
If number of combinations 'n' objects taken 'r' at a time then,
nCr = n! / ⟨ r! x (n-r)! ⟩
EXAMPLE :
1. Total number of ways to select 4 articles from a lot of 10.
= 10C4 =10!/ ⟨ 4! x (10-4)! ⟩
=10!/ ⟨ 4! x 6! ⟩
= (10 x 9 x 8 x 7 x 6! )/( 4 x 3 x 2 x 1 x 6!)
= (10 x 9 x 8 x 7)/(4 x 3 x 2 x1)
= 210
2. 60C5 = 60!/ ⟨ 5! x (60-5)! ⟩
= 60!/ ⟨ 5! x 55! ⟩
=(60 x59 x58 x57x56 x55! )/(5 x 4 x 3 x 2 x 1 x55! )
= ( 60 x59 x58 x57x56 )/ ( 5 x 4 x 3 x 2 x 1 )
= 5461512
3. A lot of 30 bulbs out of which 5 are defective, 4 times are drawn from it. Find the probability that none is defective.
Solution :
For first draw :
The number of non defective bulbs = 30-5 = 25
Hence, the probability that the first item is non defective = 25/30.
For second draw :
Total remains bulbs after first draw = 30-1 = 29 bulbs
Total remains non-defective bulb after one draw = 25-1 = 24
Hence, the probability that second item is non-defective = 24/29
Similarly,
For third draw = 23/28,
For forth draw = 22/27
Therefore, the probability that 4 times drawn are non-defective -
= 25/30 x 24/29 x 23/28 x 22/27 = 0.4616 = 46.16%
4. A lot of 30 articles contains 5 defectives. A sample of 6 selected at random from the lot. What are the respective probabilities to get 0,1,2,3,4,5 defective articles in the sample of 6.
Solution :
Here Lot Quantity = 30
No. of defective articles = 5
Hence, No. of non-defective articles = 30-5 = 25
Here,
No of ways to select 6 articles from the lot of 30 = 30C6
= 30!/( 6! x 24! )
(i). Probability of Zero defective P (0) :
Here we need to take sample from non-defective only.
Hence,
P (0) = 25C6 / 30C6
= ⟨ 25! / (6! 21! ) ⟩ / ⟨ 30! / (6! 24! ) ⟩
= ( 25! x 6! x 24! ) / ( 6! x 21! x 30! )
=(25 x24 x23 x22 x21! x 24! )/(21! x 30 x29 x 28 x 27x 26 x 25 x 24! )
= (25 x 24 x 23 x 22) / (30 x 29 x28 x27 x 26 x 25)
= 0.0007101455
(ii). Probability to get one defective article only P (1) :
" This condition possible if we get 1 defective article from 5 defective and 5 non-defective from 25 non-defective articles. "
Hence,
Probability to get 1 defective article P(1) = ( 5C1 x 25C5 ) / 30C6
= ⟨ ( 5!/(1! x 4! ) x 25!/(5! x 20!) ⟩ / ⟨ 30!/ (6! x 24!) ⟩
= 0.4474
Similarly,
(iii). Probability to get 2 defective article only P (2) :
= P (2) = ( 5C2 x 25C4 ) / 30C6
= 0.21304
(iv). Probability to get 3 defective article only P (3) :
= P (3) = ( 5C3 x 25C3 ) / 30C6
(v). Probability to get 4 defective article only P (4) :
= P (4) = ( 5C4 x 25C2 ) / 30C6
(vi). Probability to get 5 defective article only P (5) :
= P (5) = ( 5C5 x 25C1 ) / 30C6
NOTE : Here P (0)+ P(1) + P(2) + P(3) + P(4) + P(5) = 1
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