Roll one fair 6sided die. The sample space is
$\mathrm{\{1,\; 2,\; 3,\; 4,\; 5,\; 6\}}$ .
Let event
$A$ = a face is odd. Then
$\mathrm{A\; =\; \{1,\; 3,\; 5\}}$ . Let event
$B$ = a face is even.
Then
$\mathrm{B\; =\; \{2,\; 4,\; 6\}}$ .
 Find the complement of
$A$ ,
$\mathrm{A\text{'}}$ . The complement of
$A$ ,
$\mathrm{A\text{'}}$ , is
$B$ because
$A$ and
$B$ together make up the sample space.
$\text{P(A) + P(B) = P(A) + P(A') = 1}$ .
Also,
$\text{P(A)}=\frac{3}{6}$ and
$\text{P(B)}=\frac{3}{6}$
 Let event
$C$ = odd faces larger than 2. Then
$C=\{3,5\}$ . Let event
$D$ = all even
faces smaller than 5. Then
$D=\{2,4\}$ .
$\text{P(C and D)}=0$ because you cannot have an
odd and even face at the same time. Therefore,
$C$ and
$D$ are mutually exclusive
events.
 Let event
$E$ = all faces less than 5.
$E=\{1,2,3,4\}$ .
Are
$C$ and
$E$ mutually
exclusive events? (Answer yes or no.) Why or why not?
No.
$C$ =
$\mathrm{\{3,\; 5\}}$ and
$E$ =
$\mathrm{\{1,\; 2,\; 3,\; 4\}}$ .
$\mathrm{P(CANDE)}=\frac{1}{6}$ . To be mutually exclusive,
$\mathrm{P(CANDE)}$ must be 0.

$\text{Find P(CA)}$ . This is a conditional. Recall that the event
$C$ is
$\mathrm{\{3,\; 5\}}$ and event
$A$ is
$\mathrm{\{1,\; 3,\; 5\}}$ . To find
$\text{P(CA)}$ , find the probability of
$C$ using the sample space
$A$ . You
have reduced the sample space from the original sample space
$\mathrm{\{1,\; 2,\; 3,\; 4,\; 5,\; 6\}}$ to
$\mathrm{\{1,\; 3,\; 5\}}$ . So,
$\text{P(CA)}=\frac{2}{3}$
Let event
$G$ = taking a math class. Let event
$H$ = taking a science class. Then,
$\mathrm{GANDH}$ = taking a math class and a science class. Suppose
$\text{P(G)}=0.6$ ,
$\text{P(H)}=0.5$ , and
$\text{P(G AND H)}=0.3$ . Are
$G$ and
$H$ independent?
If
$G$ and
$H$ are independent, then you must show
ONE of the following:

$\text{P(GH)}=\text{P(G)}$

$\text{P(HG)}=\text{P(H)}$

$\text{P(G AND H)}=$
$\text{P(G)}\cdot \text{P(H)}$
The choice you make depends on the information you have. You could choose any of
the methods here because you have the necessary information.
Show that
$\text{P(GH)}=\text{P(G)}$ .
$\text{P(GH)}=\frac{\text{P(GANDH)}}{\text{P(H)}}=\frac{0.3}{0.5}=0.6=\text{P(G)}$
Show
$\text{P(G AND H)}=$
$\text{P(G)}\cdot \text{P(H)}$ .
$\mathrm{P(G)}\cdot \mathrm{P(H)}=0.6\cdot 0.5=0.3=\text{P(GANDH)}$
Interpretation of results
Since
$G$ and
$H$ are independent, then, knowing that a person is taking a science class does
not change the chance that he/she is taking math. (Note: IF the two events had not beenindependent  that is, IF they were dependent  then knowing that a person is taking a science
class would change the chance he/she is taking math. The next example will illustrate two events that are not independent.)
In a particular college class, 60% of the students are female. 50 % of all students in the class have long hair. 45% of the students are female and have long hair. Of the female students, 75% have long hair.
Let F be the event that the student is female. Let L be the event that the student has long hair.Are the events of being female and having long hair independent?
 The following probabilities are given in this example:

$\text{P(F )}=0.60$ ;
$\text{P(L )}=0.50$

$\text{P(F AND L)}=0.45$

$\text{P(LF)}=0.75$
If
$F$ and
$L$ are independent, then the following conditions are true. BUT if you show that any of the conditions are
not true , then
$F$ and
$L$ are
not independent .

$\text{P(LF)}=\text{P(L)}$

$\text{P(FL)}=\text{P(F)}$

$\text{P(F AND L)}=$
$\text{P(F)}\cdot \text{P(L)}$
The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(FL) yet, so you can not use the second condition.
Solution 1
Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) = (0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because P(F and L) does not equal P(F)P(L).
Solution 2
Check whether P(LF) equals P(L): We are given that P(LF) = 0.75 but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.
Interpretation of results
The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.
In a box there are 3 red cards and 5 blue cards. The red cards are
marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3,4, and 5. The cards are wellshuffled. You reach into the box (you cannot see into it) and
draw one card.
Let
$R$ = red card is drawn,
$B$ = blue card is drawn,
$E$ = evennumbered card is drawn.
The sample space
$S=\text{R1,R2,R3,B1,B2,B3,B4,B5}$ .
$S$ has 8 outcomes.

$\text{P(R)}=\frac{3}{8}$ .
$\text{P(B)}=\frac{5}{8}$ .
$\text{P(R AND B)}=0$ . (You cannot draw one card that is both
red and blue.)

$\text{P(E)}=\frac{3}{8}$ . (There are 3 evennumbered cards,
$\mathrm{R2}$ ,
$\mathrm{B2}$ , and
$\mathrm{B4}$ .)

$\text{P(EB)}=\frac{2}{5}$ . (There are 5 blue cards:
$\mathrm{B1}$ ,
$\mathrm{B2}$ ,
$\mathrm{B3}$ ,
$\mathrm{B4}$ , and
$\mathrm{B5}$ . Out of the blue cards,
there are 2 even cards:
$\mathrm{B2}$ and
$\mathrm{B4}$ .)

$\text{P(BE)}=\frac{2}{3}$ . (There are 3 evennumbered cards:
$\mathrm{R2}$ ,
$\mathrm{B2}$ , and
$\mathrm{B4}$ . Out of the
evennumbered cards, 2 are blue:
$\mathrm{B2}$ and
$\mathrm{B4}$ .)
 The events
$R$ and
$B$ are mutually exclusive because
$\text{P(R AND B)}=0$ .
 Let
$G$ = card with a number greater than 3.
$G=\{\mathrm{B4},\mathrm{B5}\}$ .
$\text{P(G)}=\frac{2}{8}$ .
Let
$H$ = blue card numbered between 1 and 4, inclusive.
$H=\{\mathrm{B1},\mathrm{B2},\mathrm{B3},\mathrm{B4}\}$ .
$\text{P(GH)}=\frac{1}{4}$ . (The only card in H that has a number greater than 3 is
$\mathrm{B4}$ .)
Since
$\frac{2}{8}=\frac{1}{4}$ ,
$\text{P(G)}=\text{P(GH)}$ which means that
$G$ and
$H$ are independent.