3.2 Independent and mutually exclusive events (modified r. bloom)  (Page 2/4)

Roll one fair 6-sided 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 $E$ = $\mathrm{\{1,\; 2,\; 3,\; 4\}}$ . $\mathrm{P(C AND E)}=\frac{1}{6}$ . To be mutually exclusive, $\mathrm{P(C AND E)}$ must be 0.

• $\text{Find P(C|A)}$ . This is a conditional. Recall that the event $C$ is $\mathrm{\{3,\; 5\}}$ and event $A$ is $\mathrm{\{1,\; 3,\; 5\}}$ . To find $\text{P(C|A)}$ , 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, $\text{P(C|A)}=\frac{2}{3}$

Let event $G$ = taking a math class. Let event $H$ = taking a science class. Then, $\mathrm{G AND H}$ = 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:

• $\text{P(G|H)}=\text{P(G)}$
• $\text{P(H|G)}=\text{P(H)}$
• $\text{P(G AND H)}=$ $\text{P(G)}\cdot \text{P(H)}$

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:
• $\text{P(F )}=0.60$ ; $\text{P(L )}=0.50$
• $\text{P(F AND L)}=0.45$
• $\text{P(L|F)}=0.75$

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 .

• $\text{P(L|F)}=\text{P(L)}$
• $\text{P(F|L)}=\text{P(F)}$
• $\text{P(F AND L)}=$ $\text{P(F)}\cdot \text{P(L)}$

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(L|F) equals P(L): We are given that P(L|F) = 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 well-shuffled. 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, $E$ = even-numbered 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 even-numbered cards, $\mathrm{R2}$ , $\mathrm{B2}$ , and $\mathrm{B4}$ .)
• $\text{P(E|B)}=\frac{2}{5}$ . (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 $\mathrm{B4}$ .)
• $\text{P(B|E)}=\frac{2}{3}$ . (There are 3 even-numbered cards: $\mathrm{R2}$ , $\mathrm{B2}$ , and $\mathrm{B4}$ . Out of the even-numbered 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. $H=\{\mathrm{B1},\mathrm{B2},\mathrm{B3},\mathrm{B4}\}$ . $\text{P(G|H)}=\frac{1}{4}$ . (The only card in H that has a number greater than 3 is $\mathrm{B4}$ .) Since $\frac{2}{8}=\frac{1}{4}$ , $\text{P(G)}=\text{P(G|H)}$ which means that $G$ and $H$ are independent.