Tree Diagrams
Aaliyah makes $70\%$ of her free throws. At the end of a close game she gets two shots. What is the chance she makes at least one? With two shots you can still list the cases in your head. With three stages, or when the chances change after each step, you need a system. That system is the tree diagram.
Building a Tree Diagram
Here is a bag with $3$ red (R) and $2$ blue (B) marbles. You draw one, note its color, put it back, and draw again. The bag is the same both times, so the second-stage branches are the same on every node.
Sometimes the second stage depends on what happened in the first. For example, you roll a die, and the roll decides which game you play next. Then the second-stage branches are different from node to node. Each node still sums to $1$.
A box contains $4$ red balls and $6$ blue balls. A ball is drawn at random, its color is noted, it is returned to the box and a second ball is drawn. Draw the…
A box contains $4$ red balls and $6$ blue balls. A ball is drawn at random, its color is noted, it is returned to the box and a second ball is drawn. Draw the corresponding tree diagram and calculate the probability of getting two blue balls.
Complete the following tree diagram that represents the flipping of a biased coin twice consecutively, knowing that the probability of getting heads is $\frac{1}{3}$.…
Complete the following tree diagram that represents the flipping of a biased coin twice consecutively, knowing that the probability of getting heads is $\frac{1}{3}$. From the tree, calculate the probability of getting heads first and then tails.
We roll a six-sided die and, then, we draw a card from a deck of $40$ cards. If an even number is rolled on the die, we win if we draw a King. If an odd number is rolled…
We roll a six-sided die and, then, we draw a card from a deck of $40$ cards. If an even number is rolled on the die, we win if we draw a King. If an odd number is rolled on the die, we win if we draw an Ace. Calculate the probability of rolling an even number on the die and also winning.
Paths: Multiply Along, Add Across
Now suppose you do not put the marble back. After the first draw the bag has one fewer marble, so the second-stage probabilities change. Take a jar with $6$ red and $4$ blue marbles, drawing two without replacement.
Back to Aaliyah's question from the start: the chance she makes at least one of her two free throws is $0.91$.
When a problem asks for an unknown count, put the unknown in the tree and write an equation.
Some problems use a $40$-card deck. It has four suits (some problems name them coins, cups, swords, and clubs) with $10$ cards in each suit: Ace, $2$ through $7$, Jack, Queen, and King. So the deck has exactly $4$ Aces and $4$ Kings, and $P(\text{King}) = \dfrac{4}{40} = \dfrac{1}{10}$ on a single draw. If you draw without replacement and the first card is of a given suit, then $9$ cards of that suit remain among $39$ cards.
In a wardrobe there are $5$ white t-shirts and $3$ black ones. Two t-shirts are chosen at random, one after the other and without returning the first one to the…
In a wardrobe there are $5$ white t-shirts and $3$ black ones. Two t-shirts are chosen at random, one after the other and without returning the first one to the wardrobe. Calculate the probability that both t-shirts are white.
A student has to take two exams on the same day. The probability of passing the first one is $0,6$. If they pass the first one, they get motivated and the probability of…
A student has to take two exams on the same day. The probability of passing the first one is $0,6$. If they pass the first one, they get motivated and the probability of passing the second one rises to $0,8$. However, if they fail the first one, they get discouraged and the probability of passing the second one drops to $0,3$. Draw the tree diagram and calculate the probability that they pass at least one of the two exams.
In an urn there are $x$ white balls and $4$ black balls. Two balls are drawn without replacement. Knowing that the probability of both balls being white is…
In an urn there are $x$ white balls and $4$ black balls. Two balls are drawn without replacement. Knowing that the probability of both balls being white is $\frac{1}{6}$, set up the corresponding equation using a tree diagram and determine the initial number of white balls $x$.
Trees That Stop Early
Not every path has the same length. In a game that ends "the moment you win" or "when you draw a red marble," a branch simply stops when the game ends. You only keep growing the branches where the game continues.
Take a bag with $2$ red and $3$ blue marbles. You draw without replacement and stop at the first red.
The same idea works for a best-of-three series, a tennis point, or any "stop when…" game. Only the branches that continue get extended.
In a tennis match, if a player's first serve goes into the box (probability $0,6$), they win the point with probability $0,7$. If the first serve misses, they hit a…
In a tennis match, if a player's first serve goes into the box (probability $0,6$), they win the point with probability $0,7$. If the first serve misses, they hit a second serve that goes in with probability $0,8$. If the second serve goes in, they win the point with probability $0,5$. If the second serve also misses, they double fault and lose the point automatically. Calculate the total probability that the player wins the point on that service.
A bag contains $5$ white balls and $4$ black balls. Balls are drawn one by one and without replacement until a black ball is drawn. What is the probability of having to…
A bag contains $5$ white balls and $4$ black balls. Balls are drawn one by one and without replacement until a black ball is drawn. What is the probability of having to make exactly three draws for the game to end?
Two e-sports teams, Alpha and Beta, play the final of a tournament in a best-of-three match (the first to win two games is the champion). Alpha is the better team, so…
Two e-sports teams, Alpha and Beta, play the final of a tournament in a best-of-three match (the first to win two games is the champion). Alpha is the better team, so their probability of winning any game is $0,6$. Games cannot end in a draw. Draw the tree diagram of the possible outcomes of the final and calculate the probability that team Beta wins the tournament.
Total Probability from a Tree
Often the first stage splits the experiment into groups, such as which machine made a part or which box you picked. To find the overall chance of an event, add up every path that ends in that event.
Chloe gets to school by bus $50\%$ of the time, walks $30\%$, and gets a ride in a car $20\%$. She is late $10\%$ of the time by bus, $5\%$ on foot, and $2\%$ by car.
The first stage doesn't have to be a "group." It can be a coin toss, a die roll, or a choice of box. If heads means "draw from urn A" and tails means "draw from urn B," the first-stage branches are $\tfrac12$ and $\tfrac12$, and the second-stage branches describe each urn.
Sometimes the first draw changes what is in the second container. Draw a counter from box 1, move it to box 2, then draw from box 2. The second-stage probabilities depend on which color moved. Build the tree with that in mind, then add the paths.
A café offers two types of menu: $60\,\%$ of the customers choose the daily menu and $40\,\%$ choose the vegetarian menu. Of those who choose the daily menu, $70\,\%$…
A café offers two types of menu: $60\,\%$ of the customers choose the daily menu and $40\,\%$ choose the vegetarian menu. Of those who choose the daily menu, $70\,\%$ order coffee for dessert. Of those who choose the vegetarian menu, $80\,\%$ order tea. Calculate the probability that a randomly chosen customer has had the daily menu and has also ordered coffee.
We have two urns. Urn A contains $3$ red balls and $2$ black balls. Urn B contains $4$ red balls and $5$ black balls. We toss a coin: if it lands on heads, we draw a…
We have two urns. Urn A contains $3$ red balls and $2$ black balls. Urn B contains $4$ red balls and $5$ black balls. We toss a coin: if it lands on heads, we draw a ball from urn A; if it lands on tails, we draw a ball from urn B. What is the probability that the drawn ball is red?
There are two boxes. Box 1 contains $4$ blue counters and $2$ red counters. Box 2 contains $3$ blue counters and $5$ red counters. A counter is drawn at random from box…
There are two boxes. Box 1 contains $4$ blue counters and $2$ red counters. Box 2 contains $3$ blue counters and $5$ red counters. A counter is drawn at random from box 1 and placed into box 2. Then, a counter is drawn from box 2. Calculate the probability that the counter drawn from box 2 is blue.
Working Backward: Conditional Probability
Chloe walks in late. How likely is it that she came by bus? Now the result is known, and you ask which first-stage branch it probably came from. The tree handles this too.
The same method answers "given" questions in game problems. If a player fell into the pit, the paths that remain are the ones that end with a fall. Divide the one path "fell on the second platform" by the sum of all falling paths.
Now you have the tools to answer any question that starts from a tree. Multiply along paths, add paths that match, and divide by the total when you are told what already happened.
In a city, $20\,\%$ of drivers are young (under 25 years old) and $80\,\%$ are adults. The probability of a young driver having an accident in a year is $0,15$, while…
In a city, $20\,\%$ of drivers are young (under 25 years old) and $80\,\%$ are adults. The probability of a young driver having an accident in a year is $0,15$, while for an adult it is $0,05$. A driver is selected at random and it turns out they have had an accident this year. Using the branches of a tree diagram, calculate the probability that this driver is young. (Express the result as an irreducible fraction).
In a board game, a player must cross a pit by jumping over three successive platforms. The probability of slipping on the first platform is $\frac{1}{10}$. If they pass…
In a board game, a player must cross a pit by jumping over three successive platforms. The probability of slipping on the first platform is $\frac{1}{10}$. If they pass the first one, the probability of slipping on the second is $\frac{1}{5}$. If they pass the second, the probability of slipping on the third is $\frac{1}{4}$. If they slip on any platform, they fall into the pit and lose. What is the probability that the player crosses the pit successfully? If we know that the player fell into the pit, what is the probability that they fell exactly on the second platform?
In a competition, an archer has a maximum of three arrows to hit the target. The probability of hitting with the first arrow is $0,4$. If they miss, they adjust their…
In a competition, an archer has a maximum of three arrows to hit the target. The probability of hitting with the first arrow is $0,4$. If they miss, they adjust their sight and the probability of hitting with the second is $0,6$. If they miss again, they get nervous and the probability of hitting with the third is $0,3$. The archer stops shooting the moment they hit the target. Calculate the probability that the archer needs to take more than one shot knowing that they finally managed to hit the target.