Two events are disjoint if they have no outcomes in common. Consider two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} that are disjoint. Can you say whether or not events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent?
Watch This
http://www.youtube.com/watch?v=tfOaGhp4L3A Brightstorm: Probability of Independent Events
Guidance
Recall that the probability of an event is the chance of it happening. Probabilities can be written as fractions or decimals between 0 and 1, or as percents between 0% and 100%. If all outcomes in an experiment have an equal chance of occurring, to find the probability of an event find the number of outcomes in the event and divide by the number of outcomes in the sample space.
\begin{align*}P(E)=\frac{\# \ of \ outcomes \ in \ E}{\# \ of \ outcomes \ in \ sample \ space}\end{align*}
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails. The sample space is \begin{align*}S=\{HH,HT,TH,TT\}\end{align*}, which contains four outcomes. Let \begin{align*}A\end{align*} be the event that heads comes up exactly once. \begin{align*}A=\{HT,TH\}\end{align*} Therefore,
\begin{align*}P(A)=\frac{2}{4}=\frac{1}{2}\end{align*}
To find the probability of a single event, all you need to do is count the number of outcomes in the event and the number of outcomes in the sample space. Probability calculations become more complex when you consider the combined probability of two or more events.
Two events are independent if one event occurring does not change the probability of the second event occurring. Two events are dependent if one event occurring causes the probability of the second event to go up or down. Two events are independent if the probability of \begin{align*}A\end{align*} and \begin{align*}B\end{align*} occurring together is the product of their individual probabilities:
\begin{align*}P(A \cap B)=P(A)P(B)\end{align*} if and only if \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent events.
In some cases it is pretty clear whether or not two events are independent. In other cases, it is not at all obvious. You can always test if two events are independent by checking to see if their probabilities satisfy the relationship above.
Example A
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails. The sample space is \begin{align*}S=\{HH,HT,TH,TT\}\end{align*}, which contains four outcomes. Let \begin{align*}C\end{align*} be the event that the first coin is a heads. Let \begin{align*}D\end{align*} be the event that the second coin is a tails.
a) List the outcomes in events \begin{align*}C\end{align*} and \begin{align*}D\end{align*}.
b) Take a guess at whether or not you think the two events are independent.
c) Find \begin{align*}P(C)\end{align*} and \begin{align*}P(D)\end{align*}.
d) Find \begin{align*}P(C \cap D)\end{align*}. Are the two events independent?
Solution:
a) \begin{align*}C=\{HH,HT\}\end{align*}. \begin{align*}D=\{HT,TT\}\end{align*}. Note that the outcomes in the two events overlap. This does NOT mean that the events are not independent!
b) If you get heads on the first coin, that shouldn't have any effect on whether you get tails for the second coin. It makes sense that the events should be independent.
c) \begin{align*}P(C)=\frac{2}{4} =\frac{1}{2}. \ P(D)=\frac{2}{4}=\frac{1}{2}\end{align*}.
d) \begin{align*}C \cap D\end{align*} is the event of getting heads first and tails second. \begin{align*}C \cap D=\{HT\}\end{align*}.
\begin{align*}P(C \cap D) &= \frac{1}{4}.\\ P(C)P(D) &= \left(\frac{1}{2} \right) \left(\frac{1}{2}\right)=\frac{1}{4}.\end{align*}
Because \begin{align*}P(C \cap D)=P(C) P(D)\end{align*}, the events are independent.
Example B
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails. The sample space is \begin{align*}S=\{HH,HT,TH,TT\}\end{align*}, which contains four outcomes. Let \begin{align*}C\end{align*} be the event that the first coin is a heads. Let \begin{align*}E\end{align*} be the event that both coins are heads.
a) List the outcomes in events \begin{align*}C\end{align*} and \begin{align*}E\end{align*}.
b) Take a guess at whether or not you think the two events are independent.
c) Find \begin{align*}P(C)\end{align*} and \begin{align*}P(E)\end{align*}.
d) Find \begin{align*}P(C \cap E)\end{align*}. Are the two events independent?
Solution:
a) \begin{align*}C=\{HH,HT\}\end{align*}. \begin{align*}E=\{HH\}\end{align*}.
b) If you get heads on the first coin, then you are more likely to end up with two heads than if you didn't know anything about the first coin. It seems like the events should NOT be independent.
c) \begin{align*}P(C)=\frac{2}{4}=\frac{1}{2}\end{align*}. \begin{align*}P(E)=\frac{1}{4}\end{align*}.
d) \begin{align*}C \cap E\end{align*} is the event of getting heads first and both heads. This is the same as the event of getting both heads, since if you got both heads then you definitely got heads first. \begin{align*}C \cap E=\{HH\}\end{align*}.
\begin{align*}P(C \cap E)&=\frac{1}{4}.\\ P(C)P(E) &= \left(\frac{1}{2} \right) \left(\frac{1}{4}\right)=\frac{1}{8}.\end{align*}
Because \begin{align*}P(C \cap E) \neq P(C)P(E)\end{align*}, the events are NOT independent.
Example C
Consider the experiment of flipping a coin two times and recording the sequence of heads and tails. The sample space is \begin{align*}S=\{HH,HT,TH,TT\}\end{align*}, which contains four outcomes. Let \begin{align*}E\end{align*} be the event that both coins are heads. Let \begin{align*}F\end{align*} be the event that both coins are tails.
a) List the outcomes in events \begin{align*}E\end{align*} and \begin{align*}F\end{align*}.
b) Take a guess at whether or not you think the two events are independent.
c) Find \begin{align*}P(E)\end{align*} and \begin{align*}P(F)\end{align*}.
d) Find \begin{align*}P(E \cap F)\end{align*}. Are the two events independent?
Solution:
a) \begin{align*}E=\{HH\}\end{align*}. \begin{align*}F\{TT\}\end{align*}.
b) If you get both heads, then you definitely didn't get both tails. It seems like the events should NOT be independent.
c) \begin{align*}P(E)=\frac{1}{4}\end{align*}. \begin{align*}P(F)=\frac{1}{4}\end{align*}.
d) \begin{align*}E \cap F\end{align*} is the event of getting two heads and two tails. This is impossible to do because these two events are disjoint. \begin{align*}E \cap F=\{\}\end{align*}.
\begin{align*}P(E \cap F) &= 0.\\ P(E) P(F) &= \left(\frac{1}{4}\right)\left(\frac{1}{4}\right)=\frac{1}{16}.\end{align*}
Because \begin{align*}P(E \cap F) \neq P(E) P(F)\end{align*}, the events are NOT independent.
Concept Problem Revisited
If \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are disjoint, then \begin{align*}A \cap B=\{\}\end{align*} and \begin{align*}P(A \cap B)=0\end{align*}.
For the two events to be independent, \begin{align*}P(A)P(B)=P(A \cap B)\end{align*} This means that \begin{align*}P(A)P(B)=0\end{align*}. By the zero product property, the only way for \begin{align*}P(A)P(B)=0\end{align*} is if \begin{align*}P(A)=0\end{align*} or \begin{align*}P(B)=0\end{align*}.
In other words, two disjoint events are independent if and only if the probability of at least one of the events is 0.
Vocabulary
An experiment is an occurrence with a result that can be observed.
An outcome of an experiment is one possible result of the experiment.
The sample space for an experiment is the set of all possible outcomes of the experiment.
An event for an experiment is a subset of the sample space containing outcomes that you are interested in (sometimes called favorable outcomes).
The complement of an event is the event that includes all outcomes in the sample space not in the original event. The symbol for complement is ′.
The union of two events is the event that includes all outcomes that are in either or both of the original events. The symbol for union is \begin{align*}\cup\end{align*}.
The intersection of two events is the event that includes all outcomes that are in both of the original events. The symbol for intersection is \begin{align*}\cap\end{align*}.
A Venn diagram is a way to visualize sample spaces, events, and outcomes.
The probability of an event is the chance of the event occurring.
Two events are independent if one event occurring does not change the probability of the second event occurring. \begin{align*}P(A \cap B)=P(A) P(B)\end{align*} if and only if \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent events.
Two events are dependent if one event occurring causes the probability of the second event to go up or down.
Two events are disjoint (mutually exclusive) if they do not have any outcomes in common.
Guided Practice
1. Consider the experiment of tossing a coin and then rolling a die. Event \begin{align*}A\end{align*} is getting a tails on the coin. Event \begin{align*}B\end{align*} is getting an even number on the die. Are the two events independent? Justify your answer using probabilities.
2. Consider the experiment of rolling a pair of dice. Event \begin{align*}C\end{align*} is a sum that is even and event \begin{align*}D\end{align*} is both numbers are greater than 4. Are the two events independent? Justify your answer using probabilities.
3. \begin{align*}P(G)=\frac{1}{3}\end{align*} and \begin{align*}P(H)=\frac{1}{2}\end{align*}. If \begin{align*}P(G \cap H)=\frac{1}{4}\end{align*}, are events \begin{align*}G\end{align*} and \begin{align*}H\end{align*} independent?
Answers:
1. The sample space for the experiment is \begin{align*}S=\{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6\}\end{align*}. Next consider the outcomes in the events. \begin{align*}A=\{T1,T2,T3,T4,T5,T6\}\end{align*}. \begin{align*}B=\{H2,H4,H6,T2,T4,T6\}\end{align*}. \begin{align*}A \cap B=\{T2, T4,T6\}\end{align*}.
 \begin{align*}P(A)=\frac{6}{12}=\frac{1}{2}\end{align*}
 \begin{align*}P(B)=\frac{6}{12}=\frac{1}{2}\end{align*}
 \begin{align*}P(A)P(B)=\left(\frac{1}{2}\right) \left(\frac{1}{2}\right)=\frac{1}{4}\end{align*}
 \begin{align*}P(A \cap B)=\frac{3}{12}=\frac{1}{4}\end{align*}
The events are independent because \begin{align*}P(A)P(B)=P(A \cap B)\end{align*}.
2. First find the sample space and the outcomes in each event:

\begin{align*}S&=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\
&(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\
&(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\end{align*}

\begin{align*}C&=\{(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5), (4,2),(4,4),\\
&(4,6), (5,1),(5,3),(5,5), (6,2),(6,4),(6,6)\}
\end{align*}
 \begin{align*}D=\{(5,5),(5,6), (6,5),(6,6)\}\end{align*}
 \begin{align*}C \cap D=\{(5,5),(6,6)\}\end{align*}
Next find the probabilities:
 \begin{align*}P(C)=\frac{18}{36}=\frac{1}{2}\end{align*}
 \begin{align*}P(D)=\frac{4}{36}=\frac{1}{9}\end{align*}
 \begin{align*}P(C)P(D)=\left(\frac{1}{2}\right)\left(\frac{1}{9}\right)=\frac{1}{18}\end{align*}
 \begin{align*}P(C \cap D)=\frac{2}{36}=\frac{1}{18}\end{align*}
The events are independent because \begin{align*}P(C)P(D)=P(C \cap D)\end{align*}.
3. Events \begin{align*}G\end{align*} and \begin{align*}H\end{align*} are independent if and only if \begin{align*}P(G)P(H)=P(G \cap H)\end{align*}.
\begin{align*}P(G)P(H) &= \left(\frac{1}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{6}\\ P(G \cap H) &= \frac{1}{4}\end{align*}
\begin{align*}P(G)P(H) \neq P(G \cap H)\end{align*}, so the events are NOT independent.
Practice
Consider the experiment of flipping a coin three times and recording the sequence of heads and tails. The sample space is \begin{align*}S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}\end{align*}, which contains eight outcomes. Let \begin{align*}A\end{align*} be the event that exactly two coins are heads. Let \begin{align*}B\end{align*} be the event that all coins are the same. Let \begin{align*}C\end{align*} be the event that at least one coin is heads. Let \begin{align*}D\end{align*} be the event that all coins are tails.
1. List the outcomes in each of the four events. Which of the two events are complements?
2. Find \begin{align*}P(A), P(B), P(C), P(D)\end{align*}.
3. Find \begin{align*}P(A \cap C)\end{align*}. Are events \begin{align*}A\end{align*} and \begin{align*}C\end{align*} independent? Explain.
4. Find \begin{align*}P(B \cap D)\end{align*}. Are events \begin{align*}B\end{align*} and \begin{align*}D\end{align*} independent? Explain.
5. Find \begin{align*}P(B \cap C)\end{align*}. Are events \begin{align*}B\end{align*} and \begin{align*}C\end{align*} independent? Explain.
6. Create two new events related to this experiment that are independent. Justify why they are independent using probabilities.
Consider the experiment of drawing a card from a deck. The sample space is the 52 cards in a standard deck. Let \begin{align*}A\end{align*} be the event that the card is red. Let \begin{align*}B\end{align*} be the event that the card is a spade. Let \begin{align*}C\end{align*} be the event that the card is a 4. Let \begin{align*}D\end{align*} be the event that the card is a diamond.
7. Describe the outcomes in each of the four events.
8. Find \begin{align*}P(A), P(B), P(C), P(D)\end{align*}.
9. Find \begin{align*}P(A \cap B)\end{align*}. Are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} independent? Explain.
10. Find \begin{align*}P(B \cap C)\end{align*}. Are events \begin{align*}B\end{align*} and \begin{align*}C\end{align*} independent? Explain.
11. Find \begin{align*}P(A \cap D)\end{align*}. Are events \begin{align*}A\end{align*} and \begin{align*}D\end{align*} independent? Explain.
12. \begin{align*}P(A)=\frac{1}{4}\end{align*} and \begin{align*}P(B)=\frac{1}{8}\end{align*}. If \begin{align*}P(A \cap B)=\frac{1}{16}\end{align*} are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} independent?
13. \begin{align*}P(A)=\frac{1}{4}\end{align*} and \begin{align*}P(B)=\frac{1}{8}\end{align*}. If \begin{align*}P(A \cap B)=\frac{1}{32}\end{align*} are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} independent?
14. What is the difference between disjoint and independent events?
15. Two events are disjoint, and both have nonzero probabilities. Can you say whether the events are independent or not?