2. MULTIPLICATIVE LAW FOR DEPENDENT EVENTS
If A and B are two dependent events in a sample spaces S , then the probability that they will both
occur is equal to the probability that one of the events will occur multiplied by the conditional
probability that other event will occur given that one event has already occurred.
P(A B) = P(A)P(B/A) Provided P(A) 0
P(B A) = P(B)P(A/B) Provided P(B) 0
Number of outcomes in (A B) = m3
Therefore, P(A) =
m1
n P(B) =
m2
n P(A B) =
m3
n
Let there are n equally likely outcomes in a sample space S.
Number of outcomes in event A = m1
S
m1
A B
A B
m2
m3
n
Number of outcome in event B = m2
Take P(A B) =
m3
n
and multiply and divide on R.H.S by m1,
we have P(A B) =
m3
n ×
m1
m1
=
m1
n ×
m3
m1
The ratio m3/m1 is the conditional probability of event B given A has occurred, that is, P(B/A).
Thus P(A B) = P(A) P(B/A)
Take P(A B) =
m3
n
and multiply and divide on R.H.S by m2, we have
P(A B) =
m3
n
×
m2
m2
=
m2
n
×
m3
m2
The ratio m3/m2 is the conditional probability of event A given B has occurred, that is, P(A/B)
Thus P(A B) = P(B) P(A/B) Amjad Ali
2
3. PROBLEM 48
A bag contains four red and three black balls. Find the probability that a red ball is drawn in
the first draw and a black ball in the second in two consecutive draws, without replacement
from the bag.
Red Black Total
4 3 7
A: Red ball on first draw
P(A) = 4/7
B: Black ball on second draw
P(B/A) = 3/6
P(A B) = P(A)P(B/A)
P(A B) = (4/7)(3/6)
P(A B) = 2/7
Red Black Total
4 3 7
1st draw 1 0
3 3 6
Amjad Ali
3
4. PROBLEM 49
Find the probability of drawing a face (pictured) card on each of two consecutive draws from
a standard pack, without replacement of the first card.
Face cards Rest cards Total
12 40 52
A: Selecting a pictured card on first draw
P(A) = 12/52
B: Selecting a pictured card on second draw
P(B/A) = 11/51
P(A B) = P(A)P(B/A)
P(A B) = (12/52)(11/51)
P(A B) = 11/221
Face Rest Total
12 40 52
1st draw 1 0
11 40 51
Amjad Ali
4
5. PROBLEM 50
A box contains 15 items, 4 of which are defective and 11 are good. Two items are selected.
What is the probability that the first is good and second is defective?
Defective Good Total
4 11 15
A: First is good
P(A) = 11/15
B: Second is defective
P(B/A) = 4/14
P(A B) = P(A)P(B/A)
P(A B) = (11/15)(4/14)
P(A B) = 22/105
Defective Good Total
4 11 15
1st item 0 1
4 10 14
Amjad Ali
5
6. PROBLEM 51
Two balls are drawn in succession without replacement from a bag containing 5 white and 8
black balls. What is the probability that both balls are black?
White Black Total
5 8 13
A: Black ball on first draw
P(A) = 8/13
B: Black ball on second draw
P(B/A) = 7/12
P(A B) = P(A)P(B/A)
P(A B) = (8/13)(7/12)
P(A B) = 14/39
White Black Total
5 8 13
1st draw 0 1
5 7 12
Amjad Ali
6
7. PROBLEM 52
Two drawing each of three balls is made from a bag containing 5 white and 8 black balls; the
balls are not being replaced before the next trial. What is the probability that the first
drawing will give 3 white balls and the second 3 black balls?
White Black Total
5 8 13
A: 3 white balls on first draw
143
5
C
C
.
C
P(A)
3
13
0
8
3
5
White Black Total
5 8 13
1st draw 3 0
2 8 10
2nd draw 0 3
B: 3 black balls on second draw
15
7
C
C
.
C
P(B/A)
3
10
3
8
0
2
P(B/A)
.
P(A)
B)
P(A
429
7
15
7
.
143
5
B)
P(A
Amjad Ali
7
8. PROBLEM 53
Show that the multiplication law P(AB) = P(A) P(B/ A), established for two events, may be
generalized to three events as follows;
P (A B C) = P (A) P (B/ A) P (C/ A B)
Let A B = K
Then A B C = K C
P(A B C) = P(K C)
= P(K ) P(C/ K)
= P(A B ) P(C/ K)
= P(A ) P(B/A ) P(C/ A B)
PROBLEM 54
Show that the multiplication law P(A B) = P(B) P(A/ B), established for two events, may be
generalized to three events as follows;
P (A B C) = P (A/ B C) P (B/ C) P ( C )
Let B C = K
Then A B C = A K
P(A B C) = P(A K)
= P(K ) P(A/ K)
= P(B C ) P(A/ K)
= P(C ) P(B/C ) P(A/ B C)
= P(A/ B C) P(B/C ) P(C ) Amjad Ali
8
9. PROBLEM 55
A farmer has a box containing 30 eggs, five of which have blood spots, he checks eggs by
taking them at random one after another from the box what is the probability that the first
two eggs have spots and third will be clear.
Spotted Clear Total
5 25 30
A: First is blood spotted
P(A) = 5/30
A: Second is blood spotted
P(B/A) = 4/29
C: Third is clear (good)
P(C/ A B) = 25/28
P(A B C) = P(A) P(B/A) P(C/A B)
P(A B C) = (5/30) (4/29) (25/28) = 25/1218
Spotted Clear Total
5 25 30
First 1 0
4 25 29
Second 1 0
3 25 28
Third 0 1
Amjad Ali
9
10. PROBLEM 56
Three cards are drawn in succession without replacement from an ordinary deck of playing
cards. Find the probability that the first card is a red ace, the second card is a ten or jack,
and the third card is greater than three but less than seven.
A: First is a red ace
P(A) = 2/52
B: Second is a ten or jack
P(B/A) = 8/51
C: Third is greater than 3 but less than 7
P(C/A B) = 12/50
P(A B C) = P(A) P(B/A) P(C/A B)
P(A B C) = (2/52) (8/51) (12/50) = 8/5525
Amjad Ali
10
