| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct distribution then calculate probability |
| Difficulty | Standard +0.3 This is a slightly-below-average A-level question. While it requires careful enumeration of outcomes and understanding of stopping conditions, the probability calculations are straightforward (powers of 1/2), and part (ii) is routine application of expectation and variance formulas with a given distribution table. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 1 } { 16 }\) | \(\frac { 11 } { 16 }\) | \(\frac { 1 } { 8 }\) | \(\frac { 1 } { 16 }\) | \(\frac { 1 } { 16 }\) |
5 A couple plan to have at least one child of each sex, after which they will have no more children. However, if they have four children of one sex, they will have no more children. You should assume that each child is equally likely to be of either sex, and that the sexes of the children are independent. The random variable $X$ represents the total number of girls the couple have.\\
(i) Show that $\mathrm { P } ( X = 1 ) = \frac { 11 } { 16 }$.
The table shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 16 }$ & $\frac { 11 } { 16 }$ & $\frac { 1 } { 8 }$ & $\frac { 1 } { 16 }$ & $\frac { 1 } { 16 }$ \\
\hline
\end{tabular}
\end{center}
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2012 Q5 [8]}}