Pre-U Pre-U 9794/3 2017 June — Question 3 8 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2017
SessionJune
Marks8
TopicDiscrete Probability Distributions
TypeConditional probability with random variables
DifficultyStandard +0.3 This is a straightforward discrete probability distribution question requiring standard techniques: summing probabilities to find k (routine algebra), calculating E(X) and E(X²) for variance (mechanical computation), and applying the conditional probability formula. All steps are textbook exercises with no novel insight required, making it slightly easier than average.
Spec2.03d Calculate conditional probability: from first principles5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

3 The probability distribution of the discrete random variable \(X\) is defined as follows. $$\mathrm { P } ( X = x ) = k ( 2 + x ) ( 5 - x ) \quad \text { for } x = 0,1,2,3,4$$
  1. Show that \(k = \frac { 1 } { 50 }\).
  2. Find the variance of \(X\).
  3. Find \(\mathrm { P } ( X = 4 \mid X > 0 )\).

Question 3(i)
- \(k \times (10 + 12 + 12 + 10 + 6) = 1\) M1 (Sum of 5 non-zero probabilities in terms of \(k\) equated to 1.)
- \(\therefore 50k = 1 \quad \therefore k = 1/50\) M1 (Shown convincingly. Depends on previous mark.)
Alternative: by verification: Sub \(k\) and all probs correct M1; Show \(\Sigma p = 1\) M1 (Shown convincingly.)
Question 3(ii)
- \(E(X) = \dfrac{0 + 12 + 24 + 30 + 24}{50} = \dfrac{9}{5}\) B1
- \(E(X^2) = \dfrac{0 + 12 + 48 + 90 + 96}{50} = \dfrac{246}{50}\) or \(4.92\) B1
- Use formula for \(\text{Var}(X)\) M1
- \(\text{Var}(X) = \dfrac{246}{50} - \left(\dfrac{9}{5}\right)^2 = \dfrac{84}{50}\) or \(1.68\) A1 (FT *their* \(E(X)\) and/or \(E(X^2)\) provided variance is positive. Accept any equivalent form.)
Question 3(iii)
- \(P(X = 4 \mid X > 0) = \dfrac{6/50}{40/50}\) M1 (Conditional probability as a ratio with either numerator or denominator correct.)
- \(= \dfrac{4}{40}\) A1 (Accept any equivalent form.)
Total: 8 marks
**Question 3(i)**
- $k \times (10 + 12 + 12 + 10 + 6) = 1$ **M1** (Sum of 5 non-zero probabilities in terms of $k$ equated to 1.)
- $\therefore 50k = 1 \quad \therefore k = 1/50$ **M1** (Shown convincingly. Depends on previous mark.)

**Alternative:** by verification: Sub $k$ and all probs correct **M1**; Show $\Sigma p = 1$ **M1** (Shown convincingly.)

**Question 3(ii)**
- $E(X) = \dfrac{0 + 12 + 24 + 30 + 24}{50} = \dfrac{9}{5}$ **B1**
- $E(X^2) = \dfrac{0 + 12 + 48 + 90 + 96}{50} = \dfrac{246}{50}$ or $4.92$ **B1**
- Use formula for $\text{Var}(X)$ **M1**
- $\text{Var}(X) = \dfrac{246}{50} - \left(\dfrac{9}{5}\right)^2 = \dfrac{84}{50}$ or $1.68$ **A1** (FT *their* $E(X)$ and/or $E(X^2)$ provided variance is positive. Accept any equivalent form.)

**Question 3(iii)**
- $P(X = 4 \mid X > 0) = \dfrac{6/50}{40/50}$ **M1** (Conditional probability as a ratio with either numerator or denominator correct.)
- $= \dfrac{4}{40}$ **A1** (Accept any equivalent form.)

**Total: 8 marks**
3 The probability distribution of the discrete random variable $X$ is defined as follows.

$$\mathrm { P } ( X = x ) = k ( 2 + x ) ( 5 - x ) \quad \text { for } x = 0,1,2,3,4$$

(i) Show that $k = \frac { 1 } { 50 }$.\\
(ii) Find the variance of $X$.\\
(iii) Find $\mathrm { P } ( X = 4 \mid X > 0 )$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2017 Q3 [8]}}