COMPLEMENT → P(A') = 1 − P(A) | OR-rule → P(A∪B) = P(A)+P(B)−P(A∩B) | MUTUALLY EXCLUSIVE → P(A∩B) = 0
Events \(A\) and \(B\) are such that \(P(A) = 0.6\), \(P(B) = 0.5\), and \(P(A \cup B) = 0.8\).
Find \(P(A \cap B)\).
⚠️ Trap: Students often forget the intersection term and add probabilities directly.
\(0.8 = 0.6 + 0.5 - P(A \cap B)\) → \(P(A \cap B) = 1.1 - 0.8 = \mathbf{0.3}\)
The probability that a student passes Mathematics is \(\frac{3}{5}\) and the probability that they pass Chemistry is \(\frac{2}{3}\). Assuming independence, find the probability that the student fails both subjects.
⚠️ Trap: "Fails both" ≠ 1 − P(passes both). You need P(fail Math) × P(fail Chem).
Since independent: P(fail both) = \(\frac{2}{5} \times \frac{1}{3} = \mathbf{\frac{2}{15}}\)
CONDITIONAL → \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) | INDEPENDENT TEST → \(P(A|B) = P(A)\) | MULTIPLY RULE → \(P(A \cap B) = P(A|B) \cdot P(B)\)
In a class, 40% of students study French, 30% study Spanish, and 15% study both. A student is chosen at random. Given that the student studies French, what is the probability they also study Spanish?
Events \(A\) and \(B\) satisfy \(P(A) = 0.4\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.2\).
Are \(A\) and \(B\) independent? And find \(P(A|B)\).
⚠️ Trap: The condition \(P(A \cap B) = P(A) \cdot P(B)\) is the independence test — many forget to verify.
\(P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4 = P(A)\) ✓
A factory has two machines. Machine A produces 60% of items and Machine B produces 40%. Machine A has a 3% defect rate; Machine B has a 5% defect rate. An item is selected at random and found to be defective. What is the probability it came from Machine B?
⚠️ Classic Bayes' Theorem problem — most common IB exam type.
P(defective) = \(0.6 \times 0.03 + 0.4 \times 0.05 = 0.018 + 0.020 = 0.038\)
P(B | defective) = \(\dfrac{0.4 \times 0.05}{0.038} = \dfrac{0.020}{0.038} \approx \mathbf{0.526}\)
ONLY A → P(A) − P(A∩B) | EXACTLY ONE → P(A∪B) − P(A∩B) | NEITHER → 1 − P(A∪B)
In a survey of 100 students: 45 play football (F), 38 play basketball (B), 20 play tennis (T). Also: \(|F \cap B| = 12\), \(|F \cap T| = 8\), \(|B \cap T| = 6\), \(|F \cap B \cap T| = 3\).
How many students play none of the three sports?
⚠️ Use inclusion-exclusion for three sets. Forgetting the triple intersection is the most common mistake.
\(|F \cup B \cup T| = 45+38+20 - 12-8-6 + 3 = 103 - 26 + 3 = 80\)
Neither = \(100 - 80 = \mathbf{16}\)
\(P(A) = 0.5\), \(P(B) = 0.4\), \(P(A' \cap B') = 0.2\).
Find \(P(A \cap B)\).
⚠️ Key: \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)\). Then use addition rule.
\(P(A \cup B) = 1 - P(A' \cap B') = 1 - 0.2 = 0.8\)
\(P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.8 = \mathbf{0.1}\)
E(X) = Σ x·P(X=x) | Var(X) = E(X²) − [E(X)]² | SD = √Var(X) | ∑P = 1 always!
A discrete random variable \(X\) has the following distribution:
\[ \begin{array}{|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 \\ \hline P(X=x) & 0.1 & 0.3 & k & 0.2 \\ \hline \end{array} \]
Find \(k\), then calculate \(E(X)\).
⚠️ Always find the missing probability first using ΣP = 1.
\(k = 1 - (0.1+0.3+0.2) = 0.4\)
\(E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = \mathbf{2.7}\)
A random variable \(X\) has \(E(X) = 3\) and \(E(X^2) = 13\).
Find the standard deviation of \(X\).
⚠️ Trap: Many calculate \(E(X^2) - E(X)^2\) but forget to take the square root at the end.
\(\text{Var}(X) = E(X^2) - [E(X)]^2 = 13 - 9 = 4\)
\(\text{SD}(X) = \sqrt{4} = \mathbf{2}\)
B(n,p) → fixed trials, 2 outcomes, constant p | P(X=r) = \(\binom{n}{r}p^r(1-p)^{n-r}\) | E(X)=np | Var=np(1−p)
A fair coin is tossed 8 times. Find the probability of getting exactly 5 heads.
\(X \sim B(8, 0.5)\)
\(P(X=5) = \binom{8}{5}(0.5)^5(0.5)^3 = 56 \times \frac{1}{256} = \dfrac{56}{256} = \mathbf{\dfrac{7}{32}}\)
The probability that a student passes a quiz is \(0.7\). The quiz is taken 6 times.
Find \(P(X \geq 5)\), i.e. the probability of passing at least 5 times.
⚠️ Trap: "At least 5" means P(X=5) + P(X=6). Don't use complement unless "at least 1".
\(X \sim B(6, 0.7)\)
\(P(X=5) = \binom{6}{5}(0.7)^5(0.3)^1 = 6 \times 0.16807 \times 0.3 \approx 0.3025\)
\(P(X=6) = (0.7)^6 \approx 0.1176\)
\(P(X \geq 5) = 0.3025 + 0.1176 \approx \mathbf{0.420}\)
\(X \sim B(n, 0.2)\). Given that \(E(X) = 4\), find \(\text{Var}(X)\) and \(\text{SD}(X)\).
⚠️ Trap: SD ≠ Var. Always take the square root for standard deviation.
\(E(X) = np = 4\) and \(p = 0.2\), so \(n = 20\).
\(\text{Var}(X) = np(1-p) = 20 \times 0.2 \times 0.8 = \mathbf{3.2}\)
\(\text{SD}(X) = \sqrt{3.2} \approx \mathbf{1.789}\)
STANDARDIZE → \(Z = \frac{X - \mu}{\sigma}\) | SYMMETRY → P(Z < −a) = P(Z > a) | INVERSE → given probability, find x value
\(X \sim N(50, 16)\). Find \(P(X > 54)\).
⚠️ Note: \(N(\mu, \sigma^2)\) means variance = 16, so \(\sigma = 4\). IB always writes the variance, not the SD.
\(\sigma = \sqrt{16} = 4\)
\(Z = \dfrac{54-50}{4} = 1\)
\(P(X > 54) = P(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = \mathbf{0.1587}\)
\(X \sim N(\mu, 25)\). It is given that \(P(X < 60) = 0.9\).
Find \(\mu\).
⚠️ Trap: Find the z-value from the table, then back-solve for μ. Many students solve it the wrong direction.
\(\sigma = 5\). \(P(Z < z) = 0.9 \Rightarrow z \approx 1.282\)
\(\dfrac{60 - \mu}{5} = 1.282 \Rightarrow \mu = 60 - 6.41 \approx \mathbf{53.59}\)
\(X \sim N(100, 225)\). Find \(P(85 < X < 115)\).
\(\sigma = 15\). \(Z_1 = \dfrac{85-100}{15} = -1\), \(Z_2 = \dfrac{115-100}{15} = 1\)
\(P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = \mathbf{0.6827}\)
(This is the famous 68% rule — within 1 SD of the mean!)
TOTAL PROB → P(B) = P(B|A)P(A) + P(B|A')P(A') | WITH/WITHOUT REPLACEMENT → changes the denominator each pick
A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement.
Find the probability that both balls are the same colour.
⚠️ Most common error: Using 8×8 in denominator instead of 8×7.
P(RR) = \(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56}\) P(BB) = \(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56}\)
P(same) = \(\dfrac{20+6}{56} = \dfrac{26}{56} = \mathbf{\dfrac{13}{28}}\)
Events \(C\) and \(D\) are mutually exclusive with \(P(C) = 0.3\) and \(P(D) = 0.4\).
Which of the following is TRUE?
⚠️ Key concept: Mutually exclusive events with non-zero probability CANNOT be independent.
Mutually exclusive → \(P(C \cap D) = 0\). Since \(P(C) \cdot P(D) = 0.12 \neq 0\), they are NOT independent.
\(P(C|D) = \dfrac{0}{0.4} = 0 \neq P(C)\), confirming dependence.
\(P(C \cup D) = 0.3 + 0.4 = \mathbf{0.7}\) ✓
Box 1 contains 4 red and 6 white balls. Box 2 contains 7 red and 3 white balls. A box is chosen at random, then a ball is drawn. Given the ball drawn is red, find the probability it came from Box 1.
P(red) = \(\frac{1}{2} \times \frac{4}{10} + \frac{1}{2} \times \frac{7}{10} = \frac{2}{10} + \frac{3.5}{10} = \frac{5.5}{10} = \frac{11}{20}\)
P(Box1 | red) = \(\dfrac{\frac{1}{2} \times \frac{4}{10}}{\frac{11}{20}} = \dfrac{\frac{4}{20}}{\frac{11}{20}} = \mathbf{\dfrac{4}{11}} \approx 0.364\)
In a large population, 30% of people are left-handed. A random sample of 200 people is taken.
Using a normal approximation to the binomial, find \(P(X \geq 70)\), where \(X\) is the number of left-handed people.
⚠️ Apply continuity correction: \(P(X \geq 70)\) becomes \(P(X > 69.5)\).
\(\mu = 200 \times 0.3 = 60\), \(\sigma^2 = 200 \times 0.3 \times 0.7 = 42\), \(\sigma \approx 6.48\)
With continuity correction: \(Z = \dfrac{69.5 - 60}{6.48} \approx 1.47\)
\(P(X \geq 70) \approx P(Z > 1.47) = 1 - 0.9292 \approx \mathbf{0.4404}\)
(Note: Without continuity correction, Z ≈ 1.543 → 0.4404 is using the rounded approach.)
The probability of hitting a target with one shot is \(\frac{1}{4}\). Shots are fired independently until the target is hit. Find the probability that the target is hit on the 4th shot.
⚠️ Geometric distribution: first success on trial \(r\) = \((1-p)^{r-1} \cdot p\). Don't use binomial here!
Geometric: \(P(X = 4) = (1-p)^{4-1} \cdot p = \left(\dfrac{3}{4}\right)^3 \times \dfrac{1}{4} = \dfrac{27}{64} \times \dfrac{1}{4} = \mathbf{\dfrac{27}{256}}\)
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