Tree diagrams in probability

Unit 11 Tree diagrams in probability

Key unit competence: Determine the probability of an event from real life situations using tree diagrams.

Key ideas

• A tree diagram shows all outcomes of an experiment done in stages. Each branch carries a probability; the branches from one point add up to 1.
• The total number of outcomes is found by multiplying the numbers of choices at each stage (e.g. 2 coins: $2\times 2=4$).
• To find the probability of a path, multiply along the branches. To combine different paths, add their probabilities.
• Mutually exclusive events cannot occur together: $P(A\text{ or }B)=P(A)+P(B)$.
• Independent events do not affect each other: $P(A\text{ and }B)=P(A)\times P(B)$.
• With replacement the probabilities stay the same at each stage; without replacement they change (the total decreases by 1).

Formulas to know

Common misconceptions

• "With two coins, 'one head and one tail' has probability $\frac{1}{4}$." No: HT and TH are two paths, so it is $\frac{1}{2}$.
• "Without replacement, the second pick has the same probabilities." No: one item has been removed.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Draw a tree diagram for tossing a coin twice and list all outcomes with their probabilities.

   Show answer

   HH, HT, TH, TT, each $\frac{1}{4}$; the four probabilities add to 1.

   Common mistake: Listing HT and TH as the same outcome.

2. Three coins are tossed. Find the probability of at least one tail.

   Show answer

   $1-P(HHH)=1-\frac{1}{8}=\frac{7}{8}$.

   Common mistake: Listing outcomes and missing some of the 8.

3. A bag contains 5 white and 3 black beads. Two beads are drawn with replacement. Find P(both white) and P(different colours).

   Show answer

   $P(WW)=\frac{25}{64}$; $P(\text{different})=2\times \frac{5}{8}\times \frac{3}{8}=\frac{30}{64}=\frac{15}{32}$.

   Common mistake: Forgetting the two orders WB and BW.

4. From 4 girls and 3 boys, two pupils are chosen at random for a task. Find the probability that both are girls.

   Show answer

   $\frac{4}{7}\times \frac{3}{6}=\frac{12}{42}=\frac{2}{7}$.

   Common mistake: Using $\frac{4}{7}\times \frac{4}{7}$ (with replacement).

5. Repeat the question without replacement.

   Show answer

   $P(WW)=\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}=\frac{5}{14}$; $P(\text{different})=\frac{5}{8}\cdot \frac{3}{7}+\frac{3}{8}\cdot \frac{5}{7}=\frac{30}{56}=\frac{15}{28}$.

   Common mistake: Keeping the denominator 8 on the second draw.

6. Show that "getting a head on a coin" and "getting a 6 on a die" are independent, and find P(both).

   Show answer

   The coin does not affect the die. $P=\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$.

   Common mistake: Calling them mutually exclusive (they can happen together).

7. The probability that Keza is late for school is 0.1 each day, independently. Find the probability that she is on time on Monday and Tuesday.

   Show answer

   $0.9\times 0.9=0.81$.

   Common mistake: Using $1-0.1\times 0.1$.

8. A team wins a match with probability 0.6, draws with 0.1 and loses with 0.3. It plays 2 matches. Find P(wins both) and P(does not lose any).

   Show answer

   $0.6^2=0.36$; not losing means W or D each time: $0.7^2=0.49$.

   Common mistake: Adding probabilities along a path.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Two coins are tossed. Find $P(HH)$.	a. $\frac{9}{14}$
   2. A box has 5 good and 3 faulty items. Two are taken without replacement. Find $P(\text{at least one faulty})$.	b. $\frac{11}{36}$
   3. A die is thrown twice. Find $P(\text{at least one six})$.	c. 0.32
   4. A bus is late with probability 0.2 each day, independently. Find $P(\text{late on exactly one of two days})$.	d. $\frac{1}{4}$

   Show answer

   1 → d ($\frac{1}{4}$) · 2 → a ($\frac{9}{14}$) · 3 → b ($\frac{11}{36}$) · 4 → c (0.32)

   • $\frac{1}{2}\times \frac{1}{2}$.
   • Step 1. $P(\text{both good})=\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}=\frac{5}{14}$.
     Step 2. $P(\text{at least one faulty})=1-\frac{5}{14}=\frac{9}{14}$.
   • Step 1. $P(\text{no six})=\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$.
     Step 2. $P(\text{at least one six})=1-\frac{25}{36}=\frac{11}{36}$.
   • Step 1. Late then on time: $0.2\times 0.8=0.16$; on time then late: $0.8\times 0.2=0.16$.
     Step 2. $0.16+0.16=0.32$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\frac{1}{4}$  ·  $\frac{3}{4}$  ·  0.384  ·  $\frac{1}{3}$
   1. A family has 2 children. Assuming boys and girls are equally likely, $P(\text{2 girls})$ is .
   2. A bus is late with probability 0.2. Over 3 days, $P(\text{late exactly once})$ is .
   3. Two coins are tossed. Find $P(\text{at least one head})$. →
   Show answer

   1: $\frac{1}{4}$ · 2: 0.384 · 3: $\frac{3}{4}$

   • GG is one of 4 equally likely outcomes.
   • $3\times 0.2\times 0.8^2=3\times 0.128$.
   • $1-P(TT)=1-\frac{1}{4}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The same spinner is spun twice. Find $P(\text{one A and one B in any order})$.	a. $\frac{3}{8}$
   2. Two dice are thrown. Find $P(\text{double six})$.	b. 0.48
   3. A coin is tossed and a die thrown. How many outcomes?	c. 12
   4. Three coins are tossed. Find $P(\text{exactly two heads})$.	d. $\frac{1}{36}$

   Show answer

   1 → b (0.48) · 2 → d ($\frac{1}{36}$) · 3 → c (12) · 4 → a ($\frac{3}{8}$)

   • $P(AB)+P(BA)=0.24+0.24$.
   • $\frac{1}{6}\times \frac{1}{6}$.
   • $2\times 6=12$.
   • Step 1. 8 equally likely outcomes; exactly two heads: HHT, HTH, THH.
     Step 2. $P=\frac{3}{8}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   0.24  ·  36  ·  18  ·  0.38
   1. Two dice are thrown. How many outcomes are there? →
   2. $P(\text{pass maths})=0.8$ and $P(\text{pass physics})=0.7$, independently. Find $P(\text{passes exactly one})$. →
   3. A spinner shows A with probability 0.6 and B with 0.4. It is spun twice. Find $P(AB)$ in that order. →
   Show answer

   1: 36 · 2: 0.38 · 3: 0.24

   • $6\times 6$.
   • Step 1. Maths only: $0.8\times 0.3=0.24$. Physics only: $0.2\times 0.7=0.14$.
     Step 2. $0.24+0.14=0.38$.
   • $0.6\times 0.4$.

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