Probability: Conditional; set algebra — ACJC 2025 H2 Math Prelim Paper 2

Part (a)

Given: \(\mathrm{P}(B\mid A) = 0.2\), \(\mathrm{P}(A\mid B) = 0.6\), \(\mathrm{P}(A'\cap B') = 0.3\).

From the conditional probability definitions: \[\begin{aligned} \mathrm{P}(B\mid A) = 0.2 &\;\Longleftrightarrow\; \mathrm{P}(A) = 5\,\mathrm{P}(A\cap B) \quad(1)\\ \mathrm{P}(A\mid B) = 0.6 &\;\Longleftrightarrow\; \mathrm{P}(B) = \tfrac{5}{3}\,\mathrm{P}(A\cap B) \quad(2) \end{aligned}\]

Since \(\mathrm{P}(A'\cap B') = 0.3\), we have \(\mathrm{P}(A\cup B) = 1 - 0.3 = 0.7\).

Using the addition formula: \[ \mathrm{P}(A\cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A\cap B) = 0.7. \] Substituting (1) and (2): \[ 0.7 = 5\,\mathrm{P}(A\cap B) + \tfrac{5}{3}\,\mathrm{P}(A\cap B) - \mathrm{P}(A\cap B) = \frac{17}{3}\,\mathrm{P}(A\cap B). \]

Alternative (Venn diagram with \(x = \mathrm{P}(A\cap B)\)):

Let \(x = \mathrm{P}(A\cap B)\). Then \(\mathrm{P}(A) = 5x\) and \(\mathrm{P}(B) = \frac{5x}{3}\), so the Venn diagram regions are:

Adding all regions: \(4x + x + \dfrac{2x}{3} + 0.3 = 1 \Rightarrow \dfrac{17x}{3} = 0.7 \Rightarrow x = \dfrac{21}{170}\).

Part (b)

Given: \(\mathrm{P}(X\cup Y\cup Z) = 0.95\); \(X\) and \(Z\) mutually exclusive; \(\mathrm{P}(Y\cap X'\cap Z') = 0.35\); \(\mathrm{P}(Y) = 0.5\).

Since \(X\cap Z = \emptyset\): \[ \mathrm{P}(Y\cap X'\cap Z') = \mathrm{P}(Y\cap X') - \mathrm{P}(Y\cap Z) = 0.35 - 0.2 = 0.15. \]

Let \(x = \mathrm{P}(X\cap Y')\). Then \(\mathrm{P}(Z\cap Y') = 1 - 0.5 - x = 0.5 - x\).

The region outside all three sets: \(1 - \mathrm{P}(X\cup Y\cup Z) = 0.05\).

We need \(x \geq 0\) and \(0.5-x \geq 0\), so \(0 \leq x \leq 0.5\).

\(\mathrm{P}(X) = x + 0.1\). When \(x = 0\): \(\mathrm{P}(X) = 0.1\) (minimum). When \(x = 0.5\): \(\mathrm{P}(X) = 0.6\) (maximum).