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Probability

Random Experiment: An experiment is said to be a random experiment, if it's out-come can't be predicted with certainty.
 
Example

If a coin is tossed, we can't say, whether head or tail will appear. So it is a random experiment.
 

 
Sample Space: The sent of all possible out-comes of an experiment is called the sample space. It is denoted by 'S' and its number of elements are n(s).
 
Example

In throwing a dice, the number that appears at top is any one of 1, 2, 3, 4, 5, 6. So here:

 

S = {1, 2, 3, 4, 5, 6} and n(s) = 6

 

Similarly in the case of a coin, S = {Head, Tail} or {H, T} and n(s) = 2.

 

The elements of the sample space are called sample point or event point.
 

 
Event: Every subset of a sample space is an event. It is denoted by 'E'.
 
Example

In throwing a dice S = {1, 2, 3, 4, 5, 6}, the appearance of an event number will be the event E = {2, 4, 6}.

 

Clearly E is a sub set of S.
 

 
Simple event; An event, consisting of a single sample point is called a simple event.
 
Example

In throwing a dice, S = {1, 2, 3, 4, 5, 6}, so each of {1}, {2}, {3}, {4}, {5} and {6} are simple events.
 

 
Compound event: A subset of the sample space, which has more than on element is called a mixed event.
 
Example

In throwing a dice, the event of appearing of odd numbers is a compound event, because E = {1, 3, 5} which has '3' elements.
 

 
Equally likely events: Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other.
 
Example

When a dice is thrown, all the six faces {1, 2, 3, 4, 5, 6} are equally likely to come up.
 

 
Exhaustive events: When every possible out come of an experiment is considered.
 
Example

A dice is thrown, cases 1, 2, 3, 4, 5, 6 form an exhaustive set of events.
 

Classical Definition of Probability

If 'S' be the sample space, then the probability of occurrence of an event 'E' is defined as:
 

 
Example
Find the probability of getting a tail in tossing of a coin.
Solution
Sample space S = {H,T} and n(s) = 2
 
Event 'E' = {T} and n(E) = 1
 
therefore P(E) = n(E)/n(S) = ½
 
Sure event: Let 'S' be a sample space. If E is a subset of or equal to S then E is called a sure event.
 
 
Example

In a throw of a dice, S = {1, 2, 3, 4, 5, 6}

 

Let E1 = Event of getting a number less than '7'.

 

So 'E1' is a sure event.

 

So, we can say, in a sure event n(E) = n(S)
 

 
Mutually exclusive or disjoint event: If two or more events can't occur simultaneously, that is no two of them can occur together.
 
Example

When a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.
 

 
Independent or mutually independent events: Two or more events are said to be independent if occurrence or non-occurrence of any of them does not affect the probability of occurrence or non-occurrence of the other event.
 
Example

When a coin is tossed twice, the event of occurrence of head in the first throw and the event of occurrence of head in the second throw are independent events.
 

 
Difference between mutually exclusive an mutually independent events: Mutually exclusiveness is used when the events are taken from the same experiment, where as independence is used when the events are taken from different experiments.
 
Complement of an event: Let 'S' be the sample for random experiment, and 'E' be an event, then complement of 'E' is denoted by E' is denoted by E'. Here E' occurs, if and only if E' doesn't occur.
 
Random Experiment: An experiment is said to be a random experiment, if its out-come can’t be predicted with certainty.
 
Example

If a coin is tossed, we can’t say, whether head or rail will appear. So it is a random experiment.

 
Sample Space: The set of all possible out-comes of an experiment is called the sample – space.
 
I a dice is thrown, the number, that appears at top is any one of 1, 2, 3, 4, 5, 6,
 
So here :
S = {1, 2, 3, 4, 5, 6,} and n(s) = 6
 
Similarly in the case of a coin, s = {H, T} and n (s) = 2.
 
The elements of the sample of the sample-space are called sample points or event points.
 
Example

If S = {H, T}, than ‘H’ and ‘T’ are sample points.
 

Important

  1. The probability of an event lies between ‘O’ and ‘1’.
     
    i.e. O< = P(E) < = 1.
  2. The probability of an impossible event is ‘O’ i.e. P(E) = O
     
    Proof: Since E has no element, = n(E) = O
     
    From definition of Probability:
     
    P(E) = n (E) / n(S) = O / n(S)
     
    = P(E) = O
  3. The probability of a sure event is 1. i.e. P(S) = 1. where ‘S’ is the sure event.
  4. If two events ‘A’ and ‘B’ are such that A ⊂ B, then P(A) < = P(B).
Question
A coin is tossed successively three times. Find the probability of getting exactly one head or two heads.
Solution
Let ‘S’ be the sample – space. Then,
 
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
= n(S) = 8
 
Let ‘E’ be the event of getting exactly one head or two heads.
 
Then:
E = {HHT, HTH, THH, TTH, THT, HTT}
= n(E) = 6
 
Therefore:
P(E) = n(E)/ n(S) = 6 / 8 = 3 / 4
  
 
Question
What is the probability, that a leap year selected at random will contain 53 Sundays?
Solution
A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.
 
The remaining 2 days may be any of the following :
  1. Sunday and Monday
  2. Monday and Tuesday
  3. Tuesday and Wednesday
  4. Wednesday and Thursday
  5. Thursday and Friday
  6. Friday and Saturday
  7. Saturday and Sunday
For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday.
 
n(S) = 7
 
n(E) = 2
 
P(E) = n(E) / n(S) = 2/7
 

Probability Using Principals of Permutation & Combination

Problems based on fundamental principal of counting and permutations and combinations:
 
Question
A bag contains ‘6’ red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability, that
  1. ‘1’ is red and ‘2’ are white,
  2. ‘2’ are blue and 1 is red,
  3. none is red.
Solution
We have to select ‘3’ balls, from 18 balls (6 + 4 + 8)
= n(S) = 18C3 = 18! / (3! × 15!) = (18 × 17 × 16)/(3 × 2 × 1) = 816
  1. Let E1 = Event of getting ‘1’ ball is red and ‘2’ are white
     
    Total number of ways = n(E1) = 6C1 × 4C2
     
    = 6!/(1! × 5!) × 4!/(2! × 2!) = 6 × 4/2 = 36
     
    P(E1) = n(E1)/n(S) = 36/816 = 3/68
  2. Let E2 = Event of getting ‘2’ balls are blue and ‘1’ is red.
     
    = Total no. of ways = n(E2) = 8C2 × 6C1
     
    = 8!/(2! × 6!) × 6!/(1! × 5!) = (8 × 7)/2 × 6/1 = 168
     
    P(E2) = 168/816 = 7/34
  3. Let E3 = Event of getting ‘3’ non – red balls. So now we have to choose all the three balls from 4 white and 8 blue balls.
     
    Total number of ways:
     
    n(E3) = 12C3 = 12! / (3! × 9!) = (12 × 11 × 10) / (3 × 2 × 1) = 220
     
    P(E3) = n(E3) / n(S) = 220/816 = 55/204
 
Remember:
  1. If ‘A’ and ‘B’ are mutually exclusive events then
     
    P(A ∩ B) = 0 or P(A and B) = 0
  2. If ‘A’ and ‘B’ by any two events, then the probability of occurrence of at least one of the events ‘A’ and ‘B’ is given by:
     
    P(A or B) = P(A) + P(B) – P(A and B)




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