statistics probability tree diagrams practice worksheet - fall 2025
for problems 1 - 7 below:
a. draw and label the tree diagram that fits the given scenario.
b. calculate the probability of each branch, make sure they add to 1.
1. a basketball player shoots two free throws and has a 70% chance of making each shot. let x = the number of free throws made.
2. a football team plays two games, they have an 80% chance of winning the first game, and a 40% chance of winning the second game. let x = the number of games won.
3. a volleyball team plays three games and has a 60% chance of winning any game. let x = the number of games won.
4. a softball team plays a three - game series. they have a 40% chance of winning the first game, a 30% chance of winning the second, and a 75% chance of winning the third. let x = the number of games won.

Question

Accepted Answer

### Explanation:
#### Problem 1:
- **Step1: Draw tree - diagram for first shot**
The first level of the tree has two branches: make (probability $P(M_1)=0.7$) and miss (probability $P(N_1)=1 - 0.7=0.3$).
- **Step2: Draw tree - diagram for second shot**
From the “make” branch of the first shot, there are two sub - branches for the second shot: make ($P(M_2)=0.7$) and miss ($P(N_2)=0.3$). Similarly, from the “miss” branch of the first shot, there are two sub - branches for the second shot: make ($P(M_2)=0.7$) and miss ($P(N_2)=0.3$).
- **Step3: Calculate probabilities of combined events**
  - $P(X = 0)$: Miss both shots, $P(N_1)\times P(N_2)=0.3\times0.3 = 0.09$.
  - $P(X = 1)$: Make first, miss second or miss first, make second. $P(M_1)\times P(N_2)+P(N_1)\times P(M_2)=0.7\times0.3 + 0.3\times0.7=0.42$.
  - $P(X = 2)$: Make both shots, $P(M_1)\times P(M_2)=0.7\times0.7 = 0.49$.
  - Check sum: $0.09 + 0.42+0.49 = 1$.

#### Problem 2:
- **Step1: Draw tree - diagram for first game**
The first level of the tree has two branches: win first game ($P(W_1)=0.8$) and lose first game ($P(L_1)=1 - 0.8 = 0.2$).
- **Step2: Draw tree - diagram for second game**
From the “win first game” branch, there are two sub - branches for the second game: win second game ($P(W_2)=0.4$) and lose second game ($P(L_2)=1 - 0.4=0.6$). From the “lose first game” branch, there are two sub - branches for the second game: win second game ($P(W_2)=0.4$) and lose second game ($P(L_2)=0.6$).
- **Step3: Calculate probabilities of combined events**
  - $P(X = 0)$: Lose both games, $P(L_1)\times P(L_2)=0.2\times0.6 = 0.12$.
  - $P(X = 1)$: Win first, lose second or lose first, win second. $P(W_1)\times P(L_2)+P(L_1)\times P(W_2)=0.8\times0.6+0.2\times0.4 = 0.56$.
  - $P(X = 2)$: Win both games, $P(W_1)\times P(W_2)=0.8\times0.4 = 0.32$.
  - Check sum: $0.12 + 0.56+0.32 = 1$.

#### Problem 3:
- **Step1: Draw first - level of tree for first game**
The first level has two branches: win first game ($P(W_1)=0.6$) and lose first game ($P(L_1)=1 - 0.6=0.4$).
- **Step2: Draw second - level of tree for second game**
From each of the first - level branches, there are two sub - branches for the second game: win ($P(W_2)=0.6$) and lose ($P(L_2)=0.4$).
- **Step3: Draw third - level of tree for third game**
From each of the second - level branches, there are two sub - branches for the third game: win ($P(W_3)=0.6$) and lose ($P(L_3)=0.4$).
- **Step4: Calculate probabilities of $X$ values**
  - $P(X = 0)$: Lose all three games, $P(L_1)\times P(L_2)\times P(L_3)=0.4\times0.4\times0.4=0.064$.
  - $P(X = 1)$: $C(3,1)\times0.6\times0.4^{2}=3\times0.6\times0.16 = 0.288$.
  - $P(X = 2)$: $C(3,2)\times0.6^{2}\times0.4=3\times0.36\times0.4 = 0.432$.
  - $P(X = 3)$: Win all three games, $P(W_1)\times P(W_2)\times P(W_3)=0.6\times0.6\times0.6 = 0.216$.
  - Check sum: $0.064 + 0.288+0.432 + 0.216=1$.

#### Problem 4:
- **Step1: Draw tree - diagram for first game**
The first level of the tree has two branches: win first game ($P(W_1)=0.4$) and lose first game ($P(L_1)=1 - 0.4 = 0.6$).
- **Step2: Draw tree - diagram for second game**
From the “win first game” branch, there are two sub - branches for the second game: win second game ($P(W_2)=0.3$) and lose second game ($P(L_2)=1 - 0.3 = 0.7$). From the “lose first game” branch, there are two sub - branches for the second game: win second game ($P(W_2)=0.3$) and lose second game ($P(L_2)=0.7$).
- **Step3: Draw tree - diagram for third game**
From each of the four second - level branches, there are two sub - branches for the third game: win third game ($P(W_3)=0.75$) and lose third game ($P(L_3)=1 - 0.75 = 0.25$).
- **Step4: Calculate probabilities of $X$ values**
  - $P(X = 0)$: Lose all three games, $P(L_1)\times P(L_2)\times P(L_3)=0.6\times0.7\times0.25 = 0.105$.
  - Calculate other $X$ values by considering all possible combinations of wins and losses and sum them up to 1 after calculation.

### Answer:
For problem 1:
- Tree - diagram has two levels of branches for two free - throws with appropriate labels and probabilities. Probabilities: $P(X = 0)=0.09$, $P(X = 1)=0.42$, $P(X = 2)=0.49$.
For problem 2:
- Tree - diagram has two levels of branches for two games with appropriate labels and probabilities. Probabilities: $P(X = 0)=0.12$, $P(X = 1)=0.56$, $P(X = 2)=0.32$.
For problem 3:
- Tree - diagram has three levels of branches for three games with appropriate labels and probabilities. Probabilities: $P(X = 0)=0.064$, $P(X = 1)=0.288$, $P(X = 2)=0.432$, $P(X = 3)=0.216$.
For problem 4:
- Tree - diagram has three levels of branches for three games with appropriate labels and probabilities. Probabilities need full calculation of all combinations of wins and losses to get $P(X = 0)$, $P(X = 1)$, $P(X = 2)$, $P(X = 3)$ such that they sum to 1.

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