Solving Basic Probability
Introduction
Introduction
Basic probability is a way of expressing information or belief that an event will occur or has occurred. In mathematics the concept has been given a correct meaning in basic probability theory that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems.
Formula used for solving basic probability
= Number of favourable out comes/Total Number of favourable outcomes
Types of Probability in Basic Concepts of Probability and Statistics:
The probability divided into three types. The basic concepts types are as follows,
Classical theory
Frequency of occurrence
Axiomatic probability theory
Basic Measurement Levels in Basic Concepts of Probability and Statistics:
Nominal
Ordinal
Interval
Ratio
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Solving Problems Based on Basic Probability:
We will use the above formula for solving basic probability.
Pro 1:In basic probability, what are the odds in favours of getting a 4 in throw of a die? What are the Events against getting a 4?
Solution:There is only one outcome favourable to the event getting a 3, the other five outcomes, namely, 1, 2, 3, 4, 5, 6 are unfavourable in basic probability. Thus,
Events in favour of getting a 4
= Number of favourable out comes / Number of outcomes
='1/6'
Events against getting a 4
= Number of unfavourable out comes/ Number of outcomes
='5/6' .
Pro 2: If the odds in favour of an event are 3 to 5, find the basic probability that it will occur.
Solution:The odds in favour of the event are '3/5'3/5. Thus,
P(A)= 1-P(A') = 1 - '3/5'
P(A) = '2/5'
The basic probability that it will occur = '2/5'
Pro 3:Sam and Alex appear for an interview for two vacancies. The basic probability of Arums selection is and that of Alexs selection is 1/6. Find the basic probability that
i) Only one of them will be selected,
ii) None of them be selected.
Solution:Let A:Sam is a selected B: Alex is selected. Then, P(A)='1/4' and P(B)='1/6'
Clearly A and not B are independent also not A are independent.
i) P(only one of them will be selected)
=P(A are not B or B and not A)
=P(A) P(not B) +P(B)P(not A)
='1/4 * ( 1 - 1/6) + 1/6 ( 1- 1/4)'
='1/4 * 5/6 + 1/6 * 3/4'
='5/24 + 3/24'
='8/24 = 1/3'
ii) P(only one of them be selected)
=P(not A and not B)
=P(not A)xP(not B)
='(1 - 1/4) * ( 1- 1/6)'
='3/4 * 5/6 = 15/24 = 5/8'
Solving Practices Problem for Basic Probability:
Pro 1:In a simultaneous toss of two coins, find the basic probability of 2 heads.
Ans: '1/4'
Pro 2:What are the odds in favours of getting a 2 in throw of a die? What are the Events against getting a 2?
Ans: '1/6' and '5/6'
More Examples
Example 1:
There are 100 boys are in the school. In those boys, there are 30 boys are playing cricket, 20 boys are playing tennis and the remaining boys are playing football. Find the probability for the following conditions?
i) Select the boys who playing the cricket.
ii) Select the boys who playing the football.
Solution
Total number of boys n(S) =100
Number of boys playing the cricket n(A)= 30
Number of boys playing tennis n(B)= 20
Number of boys playing football n(C) = 100 - (30 + 20)
= 100 - 50 = 50.
i)Assume P(A) is the probability for select the students who playing the cricket.
P(A) = '(n(A))/(n(S))'
= '30/100'
= '3/10' .
ii) Assume P(C) is the probability for select the boys who playing the football.
P(C) = '(n(C))/(n(S))'
P(C) = '50/100'
='1/2'
by: nitinp
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