Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Probability:
o

Probability of an event occurring =

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

𝒏[𝑬]

P[E] = 𝒏 [𝑺]

Notation:
o
o
o
o
o

P[A] = the probability of event A occurring
...

P[AUB] = the probability of event A or event B occurring
...

P[A I B] = the probability of event A occurring after event B has occurred
...

If event A has occurred, event B cannot occur until results have been obtained for event A
(i
...
rolling a dice can only give you one result at a time
...

If you are able to obtain results for event A and event B in the same trial, then your event
are NOT mutually exclusive
...
e
...


The Addition Rule:

▪

If two event are not mutually exclusive:
(results can be obtained for event A and event B in the same trial)
P(AUB) = P[A] + P[B] – P[AՈB]

▪

If two event are mutually exclusive:
(results for event A and event B are obtained separately)
P[AՈB] = 0
P[AUB] = P[A] + P[B] – P[AՈB]
= P[A] + P[B] + [0]
= P[A] + P[B]

Complementary Events:
If two event are complementary:
•

P[‘A] = 1 – P[A]

For events to be complementary, two conditions have to be true:
1
...

2
...

Replacement takes place
...

Replacement does NOT take place
...

P[AՈB] = P[A] X P[A I B]

Examples:

1
...
We will have to
roll the dice to obtain the result for the next event*
Multiples of 3: {3 ; 6}
P[multiples of 3] = P[3] + P[6] – P[3Ո6]
1
6
2
6
1
3

= +
=
=

1
6

− [0]

2
...
Determine the probability of selecting:
a
...
) A heart
c
...
) P[Jack] = 52
1

= 13

b
...
) *The events are not mutually exclusive because a jack and a heart can be obtained in the same event as a
jack of hearts
...
Given the sample space s = {6 ; 7 ; 8 ; 9 ; 10}
...

a
...

b
...

c
...


3

1

a
...
*
Therefore, these two event are not complementary because they are not mutually exclusive and P[A] +
P[A] ≠ 1
...
) P[ 6 ; 9 ] = 5

P [7] = 5
2

P[A] + P[B] = 5 +
3

1
5

=5
*The events are mutually exclusive because they do not share a common element and the results can
obtained in the same trial, but P[A] + P[B] ≠ 1*
Therefore, the events are not complementary
...
) P[7 ; 9] = 5

P[6 ; 8; 10] = 5
2

P[A] + P[B] = 5 +
=

5
5

3
5

=1
*the events are mutually exclusive as they do not share a common element and the results can be
obtained in the same trial and P[A] = P[B] = 1
...


5
...
In order to win, the player has to get a 6 and a head
...

a
...
) what is the probability of winning or getting another turn?

a
...

1

1

P[6] = 6

P[Head] = 2

P[AՈB] = P[6ՈHead] = P[6] x P[Head]
1

=6 ×
1

1
2

= 12
= 0
...
33% chance of winning
...
) The events are still independent; however, we now have two options
...


P[6ՈH] or P[6ՈT] = ( P[6] X P[H] ) + ( P[6] X P[T])
1
1
1
× )+(
6
2
6
1
1
( 12 ) + ( 12 )
2
12
1
6

=(
=
=
=

×

1
2

)

= 0
...
67% chance of winning or getting another turn
...
) If 3 blue balls and 7 red balls are placed in a bag, and two balls are randomly chosen without replacement,
determine the probability of choosing:
a
...

b
...

c
...

d
...


a
...
The probability of choosing a red ball is
dependent on the given fact that our blue ball was chosen first
...
2333
= 23
...
) Event A occurs before event B, and thus our events are dependent
...

P[AՈB] = P[A] + P[A I B]
=

7
10

×

3
9

7

= 30
= 0
...
33%

c
...

P[BՈR] or P[RՈB] = ( P[B] X P[B I R] ) + ( P[R] X P[R I B] )
=(
=

3
10

7
9

× )+(

7
10

×

3
9

)

7
7
+
30
30
14

= 30
=

7
15

= 0
...
67%

d
...
A blue ball is chosen for event A and a blue
ball is chosen for event B, without having replaced the first blue ball in event B
...
0667
= 6
...


6
...
45, P[B] = 0
...
615
...
) Whether event A and event B are mutually exclusive
...
) P [AՈB]
c
...


a
...
615 = 0
...
3
0
...
75
LHS ≠ RHS
Therefore, our events are not mutually exclusive
...
) P[AUB] = P[A] + P[B] – P[AՈB]
0
...
45 + 0
...
615 = 0
...
615 – 0
...
135 = - P[AՈB]
0
...
) P[AՈB] = P[A] X P[A]
0
...
45 x 0
...
135 = 0
...




---