Probability

Probability

Probability of 2 or more events

KEY WORDS TO NOTE

*Event

*Probability

*Sample space

*Possibility Space Diagram

*Outcome

*Likelihood

0-------------------------------------------1

cant happen .............................will happen

(impossible)................................(certain)

P(A) - Probability of Event A occurring

P(A') - Probability of Event A not occurring

(Think of it like a venn diagram)

Example

P(rains tomorrow) = 40% = 2/5

Therefore P(not rain tmr) = 1 - 2/5 = 3/5 or 100% - 40% = 60%

Sample Space

It refers to the set containing all the possible outcomes of an event.

Example

Tossing a die

S = { 1, 2, 3, 4, 5, 6}

n(S) = 6

There are 6 possible outcomes.

Possibility Space Diagram

This type of diagram can be used to represent all possible outcomes of 2 events occurring simultaneously.
P(A) =  $\frac{n\left(A\right)}{n\left(S\right)}$  
Example tossing 2 die.
n(S) = 36

Find the probability when
(a) P(sum=6)
(b) P(difference =3)
(c) P(product =12)

(a) P (sum= 6) = 5/36

There are 5 possible outcomes that has results when added together = 6.

[ (5,1) ( 4,2) (3,3) (2,4) (1,5) ]

(b) P (sum = 7) = 6/36 = 1/6

There are 6 possible outcomes that has results when added together = 6.

[ (6,1) (5,2) (4,3) (3,4) (2,5) (1,6)

(c) P (difference = 3) = 6/36 = 1/6

There are 6 possible outcomes that has results where the difference of the numbers = 3.

[ (4,1) (5,2) (6,3) (1,4) (2,5) (3,6) ]

(d) P (product = 12) = 4/36 = 1/9

There are 4 possible outcomes that has results where the product of the numbers = 12.

[ (6,2 ) (4,3) (3,4) (2,6) ]

ex 2

(i) Draw a possibility space diagram for tossing a die and coin at the same time.

Find the probability of the following occurring

(ii) P (H & 6)

(iii) P (T & odd no.)

(iv) P( H& Even no.)

(v) P ( H & no. > 2)

(vi) P (even no.)

n (S) =12

(ii) P (H & 6) = 1/12

[ (6,H)]

(iii) P (T & odd no.) = 3/12 = 1/4

[ (1,T) (3,T) (5,T) ]

(iv) P( H & Even no.) = 3/12 = 1/4

[ (2,H) (4,H) (6,H) ]

(v) P ( H & no. > 2) = 4/12 = 1/3

[ (3,H) (4,H) (5,H) (6,H) ]

(vi) P (even no.) = 6/12 = 1/2

[ (2, T) (4,T) (6,T) ( 2, H) (4,H) (6,H) ]

Types of Events

1. Mutually Exclusive Events
2. Non-mutually Exclusive Events
3. Independent Events

1. Mutually Exclusive Events

These are events that cannot occur at the same time.
e.g Toss a die
P (even number that is odd) = 0/6 = 0

If it was to be represented as a set / using set theory , it'll be a Disjoint Set. ( no elements in common)
1.  $A\ \cap B\ =\left\{\right\}\$  

2. Set theory  $n\ \left(A\cup B\right)=n\left(A\right)\ +\ n\left(B\right)$  

 $\therefore P\left(A\cup B\right)=\ P\left(A\right)+P\left(B\right)\$  (FORMULA)

Ex

A bag contains 10 balls : 3 white , 2 red and 5 black. What is the probability of choosing a white or a red?

n(S) = 10
P(white) = 3/10
P(red) =2/10 = 1/5
P( black) = 5/10 = 1/2

Therefore

 $P\left(W\cup R\right)=P\left(W\right)+P\left(R\right)$  
 $=\frac{3}{10}+\frac{2}{10}$  
 $=\frac{5}{10}$  
 $=\frac{1}{2}$  

2. Non-Mutually Exclusive

There are events that could happen at the same time.
Ex King of Clubs : * King and *Clubs

If it was to be represented as a set / using set theory , it'll be a Union Set. ( there are elements in common)
 $n\left(A\cup B\right)=n\left(A\right)\ +\ n\left(B\right)\ -n\left(A\cap B\right)$  

 $\therefore P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$  (formula)

Ex

A card is drawn from an ordinary pack of playing cards. (52 cards)
Find the probability that the card chosen is
1. A club or diamond
2. A club or king

P (C) = 13/52 = 1/4     P (D) = 13/52 = 1/4      P (K) = 4/52 = 1/13

1.  $P\left(C\cup D\right)=P\left(C\right)+P\left(D\right)$  

 $=\frac{13}{52}+\frac{13}{52}$  
 $=\frac{26}{52}=\frac{1}{2}$  

2.  $P\left(C\cup D\right)=P\left(C\right)+P\left(K\right)-P\left(C\cap K\right)$  
 $=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}$  
 $=\frac{4}{13}$  

3. Independent Events

If 2 events occur A and B one after the other and the P (B) occurring doesn't affect P(A) occurring then the events are independent. These events can occur simultaneously or one after the other.

Example
Toss a dice TWICE
1 toss : P (5) = 1/6
2nd toss : P(5) =1/6

 $\therefore P\left(A\cap B\right)\ =P\left(A\right)\ \times P\left(B\right)$  (formula)

Conditional Probability

This is the probability of an event A given the event B.

Formula: 
P(A/B) =  $\frac{P\left(A\cap B\right)}{P\left(B\right)}$