<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

Probability of Compound Events

Chance of an event occurring: # of successes divided by possible outcomes.

Atoms Practice
Estimated16 minsto complete
Practice Probability of Compound Events
This indicates how strong in your memory this concept is
Estimated16 minsto complete
Practice Now
Turn In
Basic Probability

Most are familiar with how flipping a coin or rolling dice works and yet probability remains one of the most counterintuitive branches of mathematics for many people. The idea that flipping a coin and getting 10 heads in a row is just as unlikely as getting the following sequence of heads and tails is hard to comprehend.


Assume a plane crashes on average once every 100 days (extremely inaccurate). Given a plane crashed today, what day in the next 100 days is the plane most likely to crash next?

Watch This

http://www.youtube.com/watch?v=YWt_u5l_jHs James Sousa: Introduction to Probability

http://www.youtube.com/watch?v=IZAMLgS5x6w James Sousa: Determining Probability


Probability is the chance of an event occurring. Simple probability is defined as the number of outcomes you are looking for (also called successes) divided by the total number of outcomes. The notation \begin{align*}P(E)\end{align*}P(E) is read “the probability of event \begin{align*}E\end{align*}E”.

\begin{align*}P(E)=\frac{\# \ successes}{\# \ possible \ outcomes}\end{align*}P(E)=# successes# possible outcomes

Probabilities can be represented with fractions, decimals, or percents. Since the number of possible outcomes is in the denominator, the probability is always between zero and one. A probability of 0 means the event will definitely not happen, while a probability of 1 means the event will definitely happen. 

\begin{align*}0 \le P(E) \le 1\end{align*}0P(E)1

The probability of something not happening is called the complement and is found by subtracting the probability from one.


You will often be looking at probabilities of two or more independent experiments. Experiments are independent when the outcome of one experiment has no effect on the outcome of the other experiment. If there are two experiments, one with outcome \begin{align*}A\end{align*}A and the other with outcome \begin{align*}B\end{align*}B, then the probability of \begin{align*}A\end{align*}A and \begin{align*}B\end{align*}B is: 

\begin{align*}P(A \ and \ B)=P(A) \cdot P(B)\end{align*}P(A and B)=P(A)P(B)

The probability of \begin{align*}A\end{align*}A or \begin{align*}B\end{align*}B is: 

\begin{align*}P(A \ or \ B)=P(A)+P(B)-P(A \ and \ B)\end{align*}P(A or B)=P(A)+P(B)P(A and B)

Example A

If you are dealt one card from a 52 card deck, what is the probability that you are dealt a heart? What is the probability that you are dealt a 3? What is the probability that you are dealt the three of hearts?

Solution: There are 13 hearts in a deck of 52 cards. \begin{align*}P(heart)=\frac{13}{52}=\frac{1}{4}\end{align*}P(heart)=1352=14

There are 4 threes in the deck of 52. \begin{align*}P(three)=\frac{4}{52}=\frac{1}{13}\end{align*}P(three)=452=113

There is only one three of hearts. \begin{align*}P(three \ and \ heart) =\frac{1}{52}\end{align*}P(three and heart)=152

Example B

Dean and his friend Randy like to play a special poker game with their friends. Dean goes home a winner 60% of the time and Randy goes home a winner 75% of the time. 

  1. What is the probability that they both win in the same night?
  2. What is the probability that Randy wins and Dean loses?
  3. What is the probability that they both lose?

Solution: First represent the information with probability symbols.

Let \begin{align*}D\end{align*}D be the event that Dean wins. Let \begin{align*}R\end{align*}R be the event that Randy wins. The complement of each probability is when Dean or Randy loses instead.

\begin{align*}P(D)=0.60, \quad P(D^C)=0.40 \end{align*}P(D)=0.60,P(DC)=0.40

\begin{align*}P(R)=0.75, \quad P(R^C)=0.25\end{align*}P(R)=0.75,P(RC)=0.25

  1. \begin{align*}P(D \ and \ R)=P(D) \cdot P(R)=0.60 \cdot 0.75=0.45\end{align*}P(D and R)=P(D)P(R)=0.600.75=0.45
  2. \begin{align*}P(R \ and \ D^C) =P(R) \cdot P(D^C)=0.75 \cdot 0.40 =0.30 \end{align*}P(R and DC)=P(R)P(DC)=0.750.40=0.30
  3. \begin{align*}P(D^C \ and \ R^C)=P(D^C) \cdot P(R^C)=0.40 \cdot 0.25=0.10\end{align*}P(DC and RC)=P(DC)P(RC)=0.400.25=0.10

Example C

If a plane crashes on average once every hundred days, what is the probability that the plane will crash in the next 100 days?

Solution: The naïve and incorrect approach would be to interpret the question as “what is the sum of the probabilities for each of the days?” Since there are 100 days and each day has a probability of 0.01 for a plane crash, then by this logic, there is a 100% chance that a plane crashes. This isn’t true because if on average the plane crashes once every hundred days, some stretches of 100 days there will be more crashes and some stretches there will be no crashes. The 100% solution does not hold.

In order to solve this question, you need to rephrase the question and ask a slightly different one that will help as an intermediate step. What is the probability that a plane does not crash in the next 100 days?

In order for this to happen it must not crash on day 1 and not crash on day 2 and not crash on day 3 etc.

The probability of the plane not crashing on any day is \begin{align*}P(no \ crash)=1-P(crash)=1-0.01=0.99\end{align*}P(no crash)=1P(crash)=10.01=0.99.

The product of each of these probabilities for the 100 days is:

\begin{align*}0.99^{100} \approx 0.366\end{align*}0.991000.366

Therefore, the probability that a plane does not crash in the next 100 days is about 36.6%. To answer the original question, the probability that a plane does crash in the next 100 days is \begin{align*}1-0.366=0.634\end{align*}10.366=0.634 or about \begin{align*}63.4 \%\end{align*}63.4%.

Concept Problem Revisited

Whether or not a plane crashes today does not matter. The probability that a plane crashes tomorrow is \begin{align*}p=0.01\end{align*}p=0.01. The probability that it crashes any day in the next 100 days is equally \begin{align*}p=0.01\end{align*}p=0.01. The key part of the question is the word “next”.

The probability that a plane does not crash on the first day and does crash on the second day is a compound probability, which means you multiply the probability of each event.

\begin{align*}P(Day \ 1 \ no \ crash \ AND \ Day \ 2 \ crash)=0.99 \cdot 0.01=0.0099\end{align*}P(Day 1 no crash AND Day 2 crash)=0.990.01=0.0099

Notice that this probability is slightly smaller than 0.01. Each successive day has a slightly smaller probability of being the next day that a plane crashes. Therefore, the day with the highest probability of a plane crashing next is tomorrow.


The probability of an event is the number of outcomes you are looking for (called successes) divided by the total number of outcomes.

The complement of an event is the event not happening.

Independent events are events where the occurrence of the first event does not impact the probability of the second event.

Guided Practice

1. Jack is a basketball player with a free throw average of 0.77. What is the probability that in a game where he has 8 shots that he makes all 8? What is the probability that he only makes 1?

2. If it has a 20% chance of raining on Tuesday, your phone has 30% chance of running out of batteries, and there is a 10% chance that you forget your wallet, what is the probability that you are in the rain without money or a phone?

3. Consider the previous question with the rain, wallet and phone. What is the probability that at least one of the three events does occur?


1. Let \begin{align*}J\end{align*}J represent the event that Jack makes the free throw shot and \begin{align*}J^C\end{align*}JC represent the event that Jack misses the shot.

\begin{align*}P(J)=0.77, \ P(J^C)=0.23\end{align*}P(J)=0.77, P(JC)=0.23

The probability that Jack makes all 8 shots is the same as Jack making one shot and making the second shot and making the third shot etc.

\begin{align*}P(J)^8=0.77^8 \approx 12.36 \%\end{align*}P(J)8=0.77812.36%

There are 8 ways that Jack could make 1 shot and miss the rest. The probability of each of these cases occurring is:

\begin{align*}P(J^C)^7 \cdot P(J)=0.23^7 \cdot 0.77\end{align*}P(JC)7P(J)=0.2370.77

Therefore, the overall probability of Jack making 1 shot and missing the rest is:

\begin{align*}0.23^7 \cdot 0.77 \cdot 8=0.0002097 =0.02097\%\end{align*}0.2370.778=0.0002097=0.02097%

2. While a pessimist may believe that all the improbable negative events will occur at the same time, the actual probability of this happening is less than one percent: 

\begin{align*}0.20 \cdot 0.30 \cdot 0.1=0.006=0.6 \%\end{align*}0.200.300.1=0.006=0.6%

3. The naïve approach would be to simply add the three probabilities together. This is incorrect. The better way to approach the problem is to ask the question: what is the probability that none of the events occur?

\begin{align*}0.8 \cdot 0.7 \cdot 0.9=0.504\end{align*}

The probability that at least one occurs is the complement of none occurring. 

\begin{align*}1-0.504=0.496=49.6 \%\end{align*}10.504=0.496=49.6%

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Probability of Compound Events.
Please wait...
Please wait...