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Addition and Multiplication Rules of Probability

Addition Rule of Probability

 
Suppose we throw a dice. What is the probability of an old number or a prime number coming up on the dice?
 
If          A is the event of getting an odd number and

            B is the event of getting a prime number,
 
Then,   Outcomes for A = 3 {1, 3, 5}
            Outcomes for B = 3 {2, 3, 5}
 
A∪B = A or B = getting an odd number or a prime number = {1, 2, 3, 5}
 
A∩B = A and B = getting an odd prime number = {3, 5}
 
 
 
Thus P (A ∪ B) = P (A) + P (B) – P(A ∩ B)
 
This result is called the Addition Rule of Probability.
 
If however the events are mutually exclusive then P(A∩B)= 0
and P(A∪B) = P(A) + P(B)
 
 

Multiplication rule of probability

 
Suppose a coin is tossed twice.
 
Let       A be the event of 'getting a head in the first toss’
            B the event of 'getting a tail in the 2nd toss'
 
Sample space = {HH, HT, TH, TT} = 4 = Total number of outcomes
 
A = {HH, HT} = 2          B = {HT, TT} = 2
 
(A∩B) = {HT} =1
 
 
The above events are independent events. If the events are dependent then
 
P (A∩B) # P (A) *  P (B)
 
 
Example : 22
 
Two dice are tossed, if E is the event of 'getting the sum of the numbers on the dice as 11' and F is the event 'getting a number other than 5 on the first dice’, find P (E∩F). Are E and F independent events?
 
Solution:
 
When two dice are cast the total number of outcomes = 6 *  6 = 36
 
Sample space S =

{(1, 1),

(1, 2),

(1, 3),

(1, 4),

(1, 5),

(1, 6)

 

(2, 1),

(2, 2),

(2, 3),

(2, 4),

(2, 5),

(2, 6)

(3, 1),

(3, 2),

(3, 3),

(3, 4),

(3, 5),

(3, 6)

(4, 1),

(4, 2),

(4, 3),

(4, 4),

(4, 5),

(4, 6)

(5, 1),

(5, 2),

(5, 3),

(5, 4),

(5, 5),

(5, 6)

(6, 1),

(6, 2),

(6, 3),

(6, 4),

(6, 5),

(6, 6)}

 

Number of outcomes favorable to E = 2 {(5, 6) (6, 5)}

 
Number of outcomes favorable to F = 30
 
Since we exclude the set {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
 
 
 
Thus, E and F are dependent events
 
In the above problem the pairs (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) are called doublets.
 

Try these problems

 
A card is drawn at random from a well shuffled deck of cards. Find the probability that the card is a

a. King or a red card

b. Club or a diamond

c. Neither a heart nor a king
 
Answer:
 
Total number of outcomes = Total number of cards
n(S) = 52
 
a.
Let A = event of getting a king
  n(A) = 4
   
  Let B = event of getting a red card

n(B) = 26

   
  Since there are 2 red kings (one of hearts and one of diamonds)

 = 2
   
 
   
  By the addition theorem of probability.

Since A and B are not mutually exclusive events

P(A or B) = P(A ∪ B) = P(A) + P(B) -
   
 
  Probability of getting a king or a red card is
   
b.
Let C = event of getting a club

 n(C) = 13

   
  D = event of getting a diamond

n(D) = 13

   
  Since getting a club or a diamond are mutually exclusive events

P(C or D) = P(C ∪ D) = P(C) +P(D)

   
  Probability of getting a club or a diamond is
   
c.
Let H = event of the card being a heart

n(H) = 13

   
 
K = event of the card being a king

n(K) = 4

   
  n  = 1 since there is a king of hearts
   
 
   
  Probability of a card being a heart or a king

by the addition theorem of probability
   
 
   
  Probability that a card is neither a heart nor a king is
   




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