Fundamentals of Probability and Statistics for Engineers

and the third axiom holds.
The definition of conditional probability given by Equation (2.24) can be
used not only to compute conditional probabilities but also to compute joint
probabilities, as the following examples show.

Example 2.9.Problem: let us reconsider Example 2.8 and ask the following
question: what is the conditional probability that the first two components are
good given that (a) the first component is good and (b) at least one of the two
is good?
Answer: the event S 1 S 2 means both are good co mponents, and S 1 S 2 is the
event that at least one of the two is good. Thus, for question (a) and in view of
Equation (2.24),

This result is expected since S 1 and S 2 are independent. Intuitively, we see that
this question is equivalent to one of co mputing P(S 2 ).
For question (b), we have

Example 2.10.Problem: in a game of cards, determine the probability of
drawing, without replacement, two aces in succession.
Answer: let A 1 be the event that the first card drawn is an ace, and similarly
for A 2. We wish to compute P(A 1 A 2 ). From Equation (2.24) we write

Now, and (there are 51 cards left and three of
them are aces). Therefore,

22 Fundamentals of Probability and Statistics for Engineers

[

[

P…S 1 S 2 jS 1 †ˆ

P…S 1 S 2 S 1 †

P…S 1 †

ˆ

P…S 1 S 2 †

P…S 1 †

ˆ

p 1 p 2

p 1

ˆp 2 :

P…S 1 S 2 jS 1 [S 2 †ˆ

P‰S 1 S 2 …S 1 [S 2 †Š

P…S 1 [S 2 †

:

Now,S 1 S 2 S 1 [S 2 )ˆS 1 S 2. Hence,

P…S 1 S 2 jS 1 [S 2 †ˆ

P…S 1 S 2 †

P…S 1 [S 2 †

ˆ

P…S 1 S 2 †

P…S 1 †‡P…S 2 †P…S 1 S 2 †

ˆ

p 1 p 2

p 1 ‡p 2 p 1 p 2

:

P…A 1 A 2 †ˆP…A 2 jA 1 †P…A 1 †: … 2 : 25 †

P A 1 )ˆ4/52 P A 2 jA 1 )ˆ3/51

P…A 1 A 2 †ˆ

3

51

4

52



ˆ

1

221

: