Probability for independent and dependent events

2. Multiplication Law for Independent Events:
If A and B are independent events having non-zero probabilities, then:
Corollary: if A and B are independent, then the following equations
are also true.
(i)
(ii)
(iii)
Example-18:
Two cards are drawn from an ordinary pack of 52 cards. The first card
is replaced before selecting the second card. Find the probability that:
(i) Both are kings
(ii) First is king and second is queen
Solution:
(i)
n (S) = 52
Let A = First Card is King
n (A) = 4
Let B = Second Card is also King
n (B) = 4
Therefore,
( ) ( ) ( )BPAPBAP =
( ) ( ) ( )BPAPBAP =
( ) ( ) ( )BPAPBAP =
( ) ( ) ( )BPAPBAP =
( ) ( )
( ) 52
4
Sn
An
AP ==
( ) ( )
( ) 52
4
Sn
Bn
BP ==
( ) ( ) ( ) 0059.0
2704
16
52
4
52
4
BPAPBAP ====

3. (ii) Let A = First Card is a King
n (A) = 4
Let B = Second Card is a Queen
n (B) = 4
Therefore,
Example-19:
The probability is 2/3 that Jawwad will pass the test and the
probability is 3/4 that Daniyal will pass the same test. Find the
probability:
(i) Both will pass the test.
(ii) Both will not pass the test.
Solution:
Let A = Jawwad will pass the test and B = Daniyal will pass the test
A = Jawwad will not pass the test and B = Daniyal will not pass the test 
;
;
(i)
(ii)
( ) ( )
( ) 52
4
Sn
An
AP ==
( ) ( )
( ) 52
4
Sn
Bn
BP ==
( ) ( ) ( ) 0059.0
2704
16
52
4
52
4
BPAPBAP ====
( )
3
2
AP = ( )
3
1
AP =
( )
4
3
BP = ( )
4
1
BP =
( ) ( ) ( ) 5.0
12
6
4
3
3
2
BPAPBAP ====
( ) ( ) ( ) 0833.0
12
1
4
1
3
1
BPAPBAP ====

4. Example-20:
Two balls are drawn from a box containing 5 red, 4 white and 6 blue
balls. The first ball is replaced before selecting the second ball. Find
the probability both are red balls.
Solution:
n (S) = 15
Let A = First Ball is red
n (A) = 5
Let B = Second ball is also red
n (B) = 5
Therefore,
( ) ( )
( ) 15
5
Sn
An
AP ==
( ) ( )
( ) 15
5
Sn
Bn
BP ==
( ) ( ) ( ) 1111.0
225
25
15
5
15
5
BPAPBAP ====

5. Multiplication Law of Dependent Events:
If A and B are two dependent events, then
Example-21:
What is the probability of getting two consecutive kings if two cards
drawn at random from an ordinary deck of 52 playing cards if the first
card is not replaced before the second card is drawn.
Solution:
n (S) = 52
Let A = First Card is King
n (A) = 4
Let B = Second Card is also King (First Card is not replaced)
n (B/A) = 3
Therefore,
( ) ( ) ( ) ( ) ( )BAPBPABPAPBAP ==
( ) ( )
( ) 52
4
Sn
An
AP ==
( ) ( )
( ) 51
3
Sn
ABn
ABP ==
( ) ( ) ( ) 0045.0
2652
12
51
3
52
4
ABPAPBAP ====

6. Example-22:
A bag contains 4 red and 6 black balls. A ball is picked at random
from the bag and not replaced before the second ball is drawn. Find
the probability that both are red balls.
Solution:
n (S) = 10
Let A = First ball is red
n (A) = 4
Let B = Second ball is also red (first ball is not replaced)
n (B/A) = 3
Therefore,
( ) ( )
( ) 10
4
Sn
An
AP ==
( ) ( )
( ) 9
3
Sn
ABn
ABP ==
( ) ( ) ( ) 1333.0
90
12
9
3
10
4
ABPAPBAP ====

7. USE Multiplicative Laws
Example23-:
A box contains 7 red and 4 yellow balls. Two balls are drawn in succession,
what is the probability that both are of same colour. If the first ball drawn is
(i) Replaced
(ii) Not Replace
Solution:
(i) P(same colour) = Both balls can be red or Both balls can be yellow.
P(same colour) = P(R1  R2) or P(Y1  Y2)
P(same colour) = P(R1)×P(R2) + P(Y1)×P(Y2)
P(same colour) =
7
11
×
7
11
+
4
11
×
4
11
P(same colour) =
49
121
+
16
121
P(same colour) =
49+16
121
P(same colour) =
65
121
= 0.537
(ii) P(same colour) = Both balls can be red or Both balls can be yellow.
P(same colour) = P(R1  R2) or P(Y1  Y2)
P(same colour) = P(R1)×P(R2/ R1)+ P(Y1)×P(Y2/ Y1)
P(same colour) =
7
11
×
6
10
+
4
11
×
3
10
P(same colour) =
42
110
+
12
110
P(same colour) =
42+12
110
P(same colour) =
54
110
= 0.4909

8. Example-24:
One bag contains 5 white balls and 4 blue balls, and a second bag
contains 4 white balls and 6 blue balls.One ball is drawn from the first
bag and placed unseen in the second bag.What is the probability that a
ball now drawn from the second bag is blue?
Solution:
P(B1) =
4
9
(Blue ball from first bag)
P(w1) =
5
9
(white ball from first bag)
If blue ball drawn from the first bag then second bag contains 4
white and 7 blue balls.
P(B2) =
7
11
(Blue ball from second bag)
If white ball drawn from the first bag then second bag contains 5
white and 6 blue balls.
P(B2) =
6
11
(Blue ball from second bag)
P(Blue from second bag) = P(B1  B2) or P(W1  B2)
P(Blue from second bag) = P(B1)×P(B2) + P(W1)×P(B2)
P(Blue from second bag) =
4
9
×
7
11
+
5
9
×
6
11
P(Blue from second bag) =
28
99
+
30
99
P(Blue from second bag) =
28+30
99
P(Blue from second bag)=
58
99
= 0.5859

9. Example-25:
A manufacturer makes writing markers. The manufacturer employs an
Inspector to check the quality of his product. The inspector tested a
random sample of markers from a large batch and calculated the
probability of any marker being defective as 0.025. Nadeem buys two
of
the markers made by the manufacturer.
(i) Calculate the probability that both markers are defective.
(ii) Calculate the probability that exactly one of the markers is
defective.
Solution:
D: A marker is defective
P(D) = 0.025
P(D/
) = 1 – 0.025 = 0.975
(i) P(Both markers are defective) = P(D  D)
P(Both markers are defective) = P(D)×P(D)
P(Both markers are defective) = 0.025×0.025
P(Both markers are defective) = 0.000625
(ii) P(Exactly one marker is defective) = P(D  D/) or P(D/  D)
P(Exactly one marker is defective) =P(D)×P(D/
) + P(D/
)×P(D)
P(Exactly one marker is defective) = 0.025×0.975 +
0.975×0.025
P(Exactly one marker is defective) =0.024375 + 0.024375
P(Exactly one marker is defective) = 0.04875

10. Example-26:
The probability that a certain type of motor will break down in the
first month of operation is 0.1. If a firm has two such motors which
are installed at the same time. Find the probability that at the end of
the first month, just one has broken down.
Solution:
1 11 ; ( ) 0.1M Motor breaks down P M= =
/ /
1 11 not ; ( ) 1 0.1 0.9M Motor breaks down P M= = − =
2 22 ; ( ) 0.1M Motor breaks down P M= =
/ /
2 22 not ; ( ) 1 0.1 0.9M Motor breaks down P M= = − =
If just one motor breaks down, then either motor 1 has broken
down and motor 2 is still working /
1 2( )P M M
Or motor 1 is still working and motor 2 has broken down
/
1 2( )P M M
Events are independent, so
P(Just one has broken down) = /
1 2( )P M M or
/
1 2( )P M M
P(Just one has broken down) =
/ /
1 2 1 2( ) ( ) ( ) ( )P M P M P M P M +
P(Just one has broken down) = 0.1×0.9 + 0.9×0.1
P(Just one has broken down) = 0.09 + 0.09 = 0.18