| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Calculate probabilities using independence |
| Difficulty | Challenging +1.2 This S4 question requires understanding of binomial distribution properties and conditional probability. Part (i) tests conceptual knowledge of independence and binomial addition, while part (ii) requires calculating a conditional probability using the definition P(A|B) = P(A∩B)/P(B) with binomial probabilities. The calculation involves multiple binomial probability terms but follows a standard approach. More challenging than typical AS-level probability due to the abstract parameter p and conditional probability with sums of random variables, but still a structured multi-part question without requiring novel insight. |
| Spec | 2.03d Calculate conditional probability: from first principles5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Successes/variables are inden with same \(p\) so B\((7+8, p)\) | B1 | or use pgf. |
| (ii) \(P(X = 2 | X + Y = 5) = \frac{P(X = 2, Y = 3)}{P(X + Y =5)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.392– (1176/3003)=56/143\) | M1, B1, B1, A1 | Numerator, Denominator |
**(i)** Successes/variables are inden with same $p$ so B$(7+8, p)$ | B1 | or use pgf.
**(ii)** $P(X = 2|X + Y = 5) = \frac{P(X = 2, Y = 3)}{P(X + Y =5)}$
$= \frac{\binom{7}{2}p^2q^5 \times \binom{8}{1}p^1q^7}{\binom{15}{5}p^5q^{10}}$
$= 0.392– (1176/3003)=56/143$ | M1, B1, B1, A1 | Numerator, Denominator1 Independent random variables $X$ and $Y$ have distributions $\mathrm { B } ( 7 , p )$ and $\mathrm { B } ( 8 , p )$ respectively.\\
(i) Explain why $X + Y \sim \mathrm {~B} ( 15 , p )$.\\
(ii) Find $\mathrm { P } ( X = 2 \mid X + Y = 5 )$.
\hfill \mbox{\textit{OCR S4 2012 Q1 [5]}}