1. **Problem 1: Drawing a tree diagram and finding probabilities for card suits** - A deck has 52 cards, 13 cards per suit (hearts, clubs, diamonds, spades). - Cards are drawn without replacement. **Tree diagram explanation:** - First draw: 4 branches (hearts, clubs, diamonds, spades), each with probability $\frac{13}{52} = \frac{1}{4}$. - Second draw: depends on first draw, 51 cards left. **Probabilities:** (i) Both cards are hearts: $$P(\text{heart}_1 \cap \text{heart}_2) = \frac{13}{52} \times \frac{12}{51} = \frac{1}{4} \times \frac{12}{51} = \frac{12}{204} = \frac{1}{17} \approx 0.0588$$ (ii) Both cards are clubs (same as hearts): $$P(\text{club}_1 \cap \text{club}_2) = \frac{13}{52} \times \frac{12}{51} = \frac{1}{17} \approx 0.0588$$ (iii) First card red (hearts or diamonds), second card black (clubs or spades): - Red cards total 26, black cards total 26. $$P(\text{red}_1 \cap \text{black}_2) = \frac{26}{52} \times \frac{26}{51} = \frac{1}{2} \times \frac{26}{51} = \frac{26}{102} = \frac{13}{51} \approx 0.2549$$ (iv) Both cards hearts given first card is heart: $$P(\text{heart}_2 | \text{heart}_1) = \frac{12}{51} \approx 0.2353$$ --- 2. **Problem 2: Tree diagram for drawing letters A, B, C, D without replacement and probabilities** - Letters: A, B, C, D. - Draw 3 letters in sequence without replacement. **Tree diagram:** - First draw: 4 options. - Second draw: 3 options. - Third draw: 2 options. **Probabilities:** (i) Spell "CAD": $$P = \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{24} \approx 0.0417$$ (ii) Spell "BAD": $$P = \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{24} \approx 0.0417$$ (iii) Spell "DAD": - Impossible because only one D tile. $$P = 0$$ (iv) Probability of not drawing letter B in 3 draws: - Letters without B: A, C, D (3 letters). - Number of ways to draw 3 letters without B is permutations of these 3 letters: $3! = 6$. - Total number of 3-letter sequences from 4 letters: $4 \times 3 \times 2 = 24$. $$P = \frac{6}{24} = \frac{1}{4} = 0.25$$ (v) Probability letters drawn in alphabetical order (A, B, C, D): - Total permutations: 24. - Only one sequence is strictly alphabetical (A then B then C or D, but only 3 draws). - For 3 draws, alphabetical sequences are those where letters are in ascending order. - Number of 3-letter ascending sequences from 4 letters is $\binom{4}{3} = 4$. - Each sequence has probability $\frac{1}{24}$. $$P = \frac{4}{24} = \frac{1}{6} \approx 0.1667$$ --- 3. **Problem 3: Venn diagram for coffee and tea preferences in 25 people** - Total people: 25 - Like coffee (C): 15 - Like tea (T): 17 - Like neither: 2 **Calculate intersection:** $$|C \cup T| = 25 - 2 = 23$$ $$|C \cup T| = |C| + |T| - |C \cap T|$$ $$23 = 15 + 17 - |C \cap T|$$ $$|C \cap T| = 15 + 17 - 23 = 9$$ **Probabilities:** (a) Like coffee: $$P(C) = \frac{15}{25} = 0.6$$ (b) Like coffee given like tea: $$P(C|T) = \frac{|C \cap T|}{|T|} = \frac{9}{17} \approx 0.5294$$ --- 4. **Problem 4: Venn diagram for coding and animation at computer camp** - Total teenagers: 100 - Learned coding: 80 - Learned animation: 42 - Each did at least one activity **Calculate both activities:** $$|C \cup A| = 100$$ $$|C \cup A| = |C| + |A| - |C \cap A|$$ $$100 = 80 + 42 - |C \cap A|$$ $$|C \cap A| = 80 + 42 - 100 = 22$$ **Probabilities:** (i) Learned coding but not animation: $$|C \setminus A| = |C| - |C \cap A| = 80 - 22 = 58$$ $$P = \frac{58}{100} = 0.58$$ (ii) Learned animation given learned coding: $$P(A|C) = \frac{|C \cap A|}{|C|} = \frac{22}{80} = 0.275$$ --- 5. **Problem 5: Probability from device preference table** | Device | School A | School B | |-----------------|----------|----------| | Smartphone | 52 | 48 | | Gaming console | 37 | 23 | | PC | 48 | 35 | **Totals:** - Total students = 52+48+37+23+48+35 = 243 (a) Probability prefers gaming console: $$P = \frac{37 + 23}{243} = \frac{60}{243} \approx 0.247$$ (b) Given gaming console, probability from School A: $$P(A|\text{console}) = \frac{37}{60} \approx 0.617$$ (c) Probability student from School A prefers PC: - Total School A students = 52 + 37 + 48 = 137 $$P = \frac{48}{137} \approx 0.350$$ (d) Probability student who prefers smartphone or console attends School B: - Smartphone or console total = (52+48) + (37+23) = 100 + 60 = 160 - School B smartphone or console = 48 + 23 = 71 $$P = \frac{71}{160} = 0.444$$