Discrete Mathematics for Computer Science

502 CHAPTER 8 Discrete Probability

On the other hand,

P (E1) = Ell (3)=

and

P(E 2 ) (Di E 21 = (26) =

Hence,

1

P(EI) • P(E 2 ) - = P(E 1 x E 2 D)

4

Example 5. Suppose we flip a coin once and roll a die twice. Formulate the event

E = "heads on the coin and an even number on the second roll"

as a cross product of sets in a cross product sample space. Calculate the probability of the

event.

Solution. We choose sample spaces

Q2 1 ={H,TJ and 2 = 023 = {1, 2, 3, 4,5, 61

and assign each a uniform probability density. Flipping the coin and rolling the die

twice is modeled by the sample space QŽ1 x Q2 X 03. Let E 1 = {H}, E 2 = Q22, and

E3 = 12, 4, 6). Since the outcome of the first roll is not specified, all possibilities must

be taken into account. Hence,

E = E 1 x Q22 x E 3

and

P(E) = P(E 1 ) • P(Q 2 ) -P(E 3 ) = (1) =

8.3.5 Two Ways of Viewing Events

Let

S= 1 X ... X×

be the cross product of sample spaces 2 1, Q2. .2n,. This section looks at the relation-

ship between events in Q^2 i and events in Q.

Suppose we want to specify an event in the cross product sample space QŽ just by

specifying what happens in one of its component sample spaces £Ži (we do not care what

happens in the other component sample spaces). We can reformulate a nonempty event

Ei in Qi as a corresponding event E* in Q1 X 22 X ... x ,2 by setting E* equal to all

n-tuples in which the ith component belongs to Ei.

For example, suppose we flip a penny and roll a die. Then, n = 2, 21 = {H, T}, and

02 = {1, 2, 3, 4, 5, 6}. Getting an even number on the die is the event E 2 = {2, 4, 61 C Q2.