Tree Diagrams, Venn Diagrams, Set Notation & Conditional Probability

1

The and Rule (Multiply) & OR Rule (Add)

1. For two independent events A and B, the probability of both happening is $P(A \text{ and } B) = P(A) \times P(B)$ (multiply along the branches of a tree diagram). 2. For two mutually exclusive events (events that cannot happen at the same time), the probability of A OR B happening is $P(A \text{ or } B) = P(A) + P(B)$ (add the end results of different branch paths).

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2

Tree diagrams without replacement (conditional)

When two items are chosen at random from a group of $N$ items without replacement, the probabilities on the second set of branches depend on what happened first. Always check that every pair (or trio) of branches starting from the same node adds up to 1, and that the denominator on the second pick is $(N - 1)$.

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3

Venn Diagrams & Set Notation ($\xi, A \cap B, A \cup B, A'$)

Always fill in a Venn diagram starting from the very centre intersection ($A \cap B$) and work outwards, subtracting the overlap from the total of each circle so you never double-count $\xi$ is the universal set (every element inside the rectangle); $A \cap B$ is 'A and B'; $A \cup B$ is 'A or B or both'; $A'$ is 'not A'; $(A \cup B)'$ is the region outside both circles.

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4

Conditional Probability from Venn Diagrams ($P(A | B)$)

The probability of event A occurring given that event B has already occurred is written $P(A | B)$ ('probability of A given B'). Formula: $P(A | B) = \frac{\text{Number in }(A \cap B)}{\text{Total number in } B}$. Only divide by the total inside circle B, not the universal total.

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5

Algebraic Probability without Replacement (Higher Tier)

In Grade 9 questions, there are $n$ items in a bag (e.g., 6 red counters and $x$ blue counters, or $n$ sweets in total of which 4 are orange). Write the branch probabilities as algebraic fractions: on the first pick the denominator is $n$, and on the second pick without replacement the denominator is $(n - 1)$! Multiply the algebraic fractions, set equal to the given probability, cross-multiply to remove fractions, and solve the resulting quadratic equation.

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6

Expected Frequency (Relative Frequency)

If the probability of an event occurring in a single trial is $p$ and the experiment is repeated $N$ times, the expected number of times the event will occur is: $\text{Expected Frequency} = N \times p$. Increasing the number of trials $N$ makes the experimental relative frequency closer to the true theoretical probability.

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1. Q1

   A box contains 4 Milk chocolates, 5 Dark chocolates, and 3 White chocolates. Sarah takes two chocolates at random from the box and eats them (without replacement). Calculate the probability that the two chocolates Sarah eats are of different types. [4 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Total number of chocolates initially = 4 + 5 + 3 = 12 (so the first denominator is 12 and the second denominator without replacement is 11) [1 mark]. 2. Use the complement shortcut: P(Two different types) = 1 - P(Both the same type) = 1 - [P(MM) + P(DD) + P(WW)] [1 mark]. 3. Calculate P(Same type): P(MM) = (4/12) x (3/11) = 12/132; P(DD) = (5/12) x (4/11) = 20/132; P(WW) = (3/12) x (2/11) = 6/132 -> Total P(Same) = (12 + 20 + 6) / 132 = 38 / 132 [1 mark]. 4. Subtract from 1: P(Different types) = 1 - (38 / 132) = 94 / 132 (or 47 / 66) [1 mark].

2. Q2

   There are n counters in a bag. 5 of the counters are red and the rest are blue. Two counters are taken at random from the bag without replacement. The probability that both counters are red is 5/33. Show that n² - n - 132 = 0, and hence find the total number of counters in the bag initially. (Higher Tier) [5 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Write down the probability of picking a red counter on the 1st pick (5 / n) and a red counter on the 2nd pick without replacement (4 / (n - 1)) [1 mark]. 2. Multiply the two probabilities and set equal to 5/33: (5 / n) x (4 / (n - 1)) = 5 / 33 -> 20 / (n(n - 1)) = 5 / 33 [1 mark]. 3. Cross-multiply to remove fractions: 20 x 33 = 5n(n - 1) -> 660 = 5n² - 5n [1 mark]. 4. Divide every term by 5 and rearrange to equal zero: 132 = n² - n -> n² - n - 132 = 0 (as required) [1 mark]. 5. Factorise and solve for positive n: (n - 12)(n + 11) = 0 -> since the number of counters n must be positive, n = 12 counters [1 mark].

3. Q3

   In a group of 80 sixth-form students, 42 study Biology (B), 36 study Chemistry (C), and 19 study neither Biology nor Chemistry. (a) Work out the number of students in the set B ∩ C. (b) One of the students who studies Biology is chosen at random. Write down the probability that this student does not study Chemistry (i.e., is in B ∩ C'). [4 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. (a) Subtract the 19 students who study neither from 80 to find n(B ∪ C): 80 - 19 = 61 students study Biology, Chemistry, or both [1 mark]. 2. Find the intersection n(B ∩ C) by subtracting 61 from the sum of the two circles: n(B ∩ C) = (42 + 36) - 61 = 78 - 61 = 17 students [1 mark]. 3. (b) Find the number of students who study Biology only (B ∩ C'): 42 - 17 = 25 students [1 mark]. 4. Since the student is chosen from those who study Biology (total 42), the conditional probability is 25 / 42 [1 mark].

4. Q4

   Bag A contains 3 red balls and 7 yellow balls. Bag B contains 6 red balls and 4 yellow balls. A ball is taken at random from Bag A and placed into Bag B. A ball is then taken at random from Bag B. Calculate the probability that the ball taken from Bag B is red. (Higher Tier) [4 marks]

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Consider Case 1 (Red moved from A to B): P(Red from A) = 3/10; Bag B now has 7 red and 4 yellow balls (total 11), so P(Red from B | Red from A) = 7/11 -> P(Red, Red) = (3/10) x (7/11) = 21/110 [1 mark]. 2. Consider Case 2 (Yellow moved from A to B): P(Yellow from A) = 7/10; Bag B now has 6 red and 5 yellow balls (total 11), so P(Red from B | Yellow from A) = 6/11 [1 mark]. 3. Multiply for Case 2: P(Yellow, Red) = (7/10) x (6/11) = 42/110 [1 mark]. 4. Add the two mutually exclusive cases together: P(Red from B) = 21/110 + 42/110 = 63/110 [1 mark].

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