June 2024 Paper 2 Q14
14 For a certain value of the constant \(p\), the random variable \(X\) has the probability distribution given in the table.
| \(x\) | 1 | 2 | 3 | 4 |
| \(\mathrm{P}(X = x)\) | \(p\) | \(\frac{1}{6}p\) | \(p^2\) | \(\frac{1}{2}\) |
Two independent values, \(X_1\) and \(X_2\), of \(X\) are found.
Determine \(\mathrm{P}(X_2 = 2X_1 \mid X_2 \gt X_1)\). [8]
| Scheme | Marks | AO | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(p + \frac{1}{6}p + p^2 + \frac{1}{2} = 1\) | M1* | 3.1a | ||||||||||
| \(p^2 + \frac{7}{6}p - \frac{1}{2} = 0\) or \(6p^2 + 7p - 3 = 0\) oe | M1 dep* | 1.1 | ||||||||||
| \(p = \frac{1}{3}\) | A1 | 1.1 | ||||||||||
| \(p = -\frac{3}{2}\) invalid | B1 | 2.3 | ||||||||||
(Hence probabilities are):
| A1 | 1.1 | ||||||||||
| \(\left(\dfrac{\mathrm{P}(X_2 = 2X_1 \cap X_2 \gt X_1)}{\mathrm{P}(X_2 \gt X_1)} = \dfrac{\mathrm{P}(X_2 = 2X_1)}{\mathrm{P}(X_2 \gt X_1)}\right)\) Any of: \(p \cdot \frac{p}{6} + \frac{p}{6} \cdot \frac{1}{2} = \frac{1}{3} \times \frac{1}{18} + \frac{1}{18} \times \frac{1}{2}\) or \(p \cdot \frac{p}{6} + p \cdot p^2 + p \cdot \frac{1}{2} + \frac{p}{6} \cdot p^2 + \frac{p}{6} \cdot \frac{1}{2} + p^2 \cdot \frac{1}{2}\) \(= \frac{1}{3} \times \frac{1}{18} + \frac{1}{3} \times \frac{1}{9} + \frac{1}{3} \times \frac{1}{2} + \frac{1}{18} \times \frac{1}{9} + \frac{1}{18} \times \frac{1}{2} + \frac{1}{9} \times \frac{1}{2}\) | M1 | 2.1 | ||||||||||
| \(\dfrac{p \cdot \frac{p}{6} + \frac{p}{6} \cdot \frac{1}{2}}{p \cdot \frac{p}{6} + p \cdot p^2 + p \cdot \frac{1}{2} + \frac{p}{6} \cdot p^2 + \frac{p}{6} \cdot \frac{1}{2} + p^2 \cdot \frac{1}{2}}\) \(= \dfrac{\frac{1}{3} \times \frac{1}{18} + \frac{1}{18} \times \frac{1}{2}}{\frac{1}{3} \times \frac{1}{18} + \frac{1}{3} \times \frac{1}{9} + \frac{1}{3} \times \frac{1}{2} + \frac{1}{18} \times \frac{1}{9} + \frac{1}{18} \times \frac{1}{2} + \frac{1}{9} \times \frac{1}{2}}\) oe \(\left(= \dfrac{\frac{5}{108}}{\frac{101}{324}}\ \text{or}\ \dfrac{0.0463}{0.3117}\right)\) | M1 | 1.1 | ||||||||||
| \(= \frac{15}{101}\) or 0.149 (3 sf) | A1 | 1.1 | ||||||||||
| [8] |
Notes
M1*: Forming this equation in \(p\), must be fully correct with \(= 1\) soi
M1 dep*: Rearrange their equation to solvable form \(ap^2 + bp + c = 0\) and attempt to solve (may be implied by one or both correct roots)
B1: For sight of \(p = -\frac{3}{2}\) oe provided this root not used in subsequent working. Condone “the other root is negative” or “\(p \gt 0\)”
A1: These values are likely to be seen in, and may be implied by, subsequent working. Note that a correct numerical denominator or final answer also implies this mark.
M1: Either numerator or denominator attempted (correct form with 2 terms in the numerator or 6 terms in the denominator). FT their probabilities.
May be implied by any of:
• a correct expression for one of the numerator or denominator either in \(p\) or with their probabilities
• a correct final answer
M1: Division attempted with a 2-term numerator and 6-term denominator soi either in \(p\) or with their probabilities (may see numerator and denominator computed separately and then an attempt to divide)
A1: oe, need not be simplified