## Rules of Probability

#### Addition and Multiplication Rules 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

Total number of outcomes = Total number of cards

n(S) = 52

1. 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
2. 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
3. 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 