| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.2 This is a straightforward conditional probability question requiring only reading values from a two-way table and applying P(A|B) = P(A∩B)/P(B). Part (i) is verification by addition, parts (ii)-(iv) are direct one-step calculations with no conceptual difficulty beyond understanding the definition of conditional probability. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Weather Forecast | \multirow{2}{*}{Total} | ||||||
| \cline { 3 - 6 } \multicolumn{2}{|c|}{} | Sunny | Cloudy | Wet | ||||
\multirow{3}{*}{
| Sunny | 55 | 12 | 7 | 74 | ||
| \cline { 2 - 6 } | Cloudy | 17 | 128 | 29 | 174 | ||
| \cline { 2 - 6 } | Wet | 3 | 33 | 81 | 117 | ||
| Total | 75 | 173 | 117 | 365 | |||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{Correct forecast}) = \dfrac{55+128+81}{365} = \dfrac{264}{365}\) | M1, A1 | Numerator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{Correct forecast} \mid \text{sunny}) = \dfrac{55}{75} = 0.733\) | M1, A1 | Denominator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{Correct forecast} \mid \text{wet}) = \dfrac{81}{117} = 0.692\) | M1, A1 | Denominator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{Cloudy} \mid \text{correct forecast}) = \dfrac{128}{264} = 0.485\) | M1, A1 | Denominator |
## Question 6:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{Correct forecast}) = \dfrac{55+128+81}{365} = \dfrac{264}{365}$ | M1, A1 | Numerator |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{Correct forecast} \mid \text{sunny}) = \dfrac{55}{75} = 0.733$ | M1, A1 | Denominator |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{Correct forecast} \mid \text{wet}) = \dfrac{81}{117} = 0.692$ | M1, A1 | Denominator |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{Cloudy} \mid \text{correct forecast}) = \dfrac{128}{264} = 0.485$ | M1, A1 | Denominator |
---6 An amateur weather forecaster describes each day as either sunny, cloudy or wet. He keeps a record each day of his forecast and of the actual weather. His results for one particular year are given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
\multicolumn{2}{|c|}{} & \multicolumn{3}{|c|}{Weather Forecast} & \multirow{2}{*}{Total} \\
\cline { 3 - 6 }
\multicolumn{2}{|c|}{} & Sunny & Cloudy & Wet & \\
\hline
\multirow{3}{*}{\begin{tabular}{ c }
Actual \\
Weather \\
\end{tabular}} & Sunny & 55 & 12 & 7 & 74 \\
\cline { 2 - 6 }
& Cloudy & 17 & 128 & 29 & 174 \\
\cline { 2 - 6 }
& Wet & 3 & 33 & 81 & 117 \\
\hline
\multicolumn{2}{|c|}{Total} & 75 & 173 & 117 & 365 \\
\hline
\end{tabular}
\end{center}
A day is selected at random from that year.\\
(i) Show that the probability that the forecast is correct is $\frac { 264 } { 365 }$.
Find the probability that\\
(ii) the forecast is correct, given that the forecast is sunny,\\
(iii) the forecast is correct, given that the weather is wet,\\
(iv) the weather is cloudy, given that the forecast is correct.
\hfill \mbox{\textit{OCR MEI S1 2005 Q6 [8]}}