Edexcel S1 2002 January — Question 4 10 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2002
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndependent Events
TypeTest independence using definition
DifficultyModerate -0.8 This is a straightforward probability question testing basic set operations and independence definition. Parts (a)-(c) use standard Venn diagram or complement rules with given probabilities. Part (d) tests the definition P(A∩B) = P(A)×P(B), requiring only arithmetic comparison. All techniques are routine S1 content with no problem-solving insight needed.
Spec2.03a Mutually exclusive and independent events

A contractor bids for two building projects. He estimates that the probability of winning the first project is 0.5, the probability of winning the second is 0.3 and the probability of winning both projects is 0.2.
  1. Find the probability that he does not win either project. [3]
  2. Find the probability that he wins exactly one project. [2]
  3. Given that he does not win the first project, find the probability that he wins the second. [2]
  4. By calculation, determine whether or not winning the first contract and winning the second contract are independent events. [3]

Part (a)
AnswerMarks Guidance
Venn diagramM1 0.3, 0.2, 0.1
\(P(\text{does not win either}) = 0.4\)A1 (3)
Part (b)
AnswerMarks
\(P(\text{wins exactly one}) = 0.3 + 0.1 = 0.4\)M1
A1/(2)
Part (c)
AnswerMarks Guidance
\(P(B_1B_1') = \frac{P(B_2 \cap B_1')}{P(B_1')} = \frac{0.1}{0.5} = 0.2\) M1
A1 (2)
Part (d)
AnswerMarks Guidance
For independence: \(P(B_1 \cap B_2) = P(B_1) \times P(B_2)\)M1
\(P(B_1 \cap B_2) = 0.2\); \(P(B_1) \times P(B_2) = 0.15\)0.2; 0.15 A1
LHS \(\neq\) RHS \(\Rightarrow\) events not independent A1 (3)
NB: Accept alternate correct solutions
## Part (a)
Venn diagram | M1 | 0.3, 0.2, 0.1
$P(\text{does not win either}) = 0.4$ | A1 (3) | 

## Part (b)
$P(\text{wins exactly one}) = 0.3 + 0.1 = 0.4$ | M1 | 
 | A1/(2) | 

## Part (c)
$P(B_1 | B_1') = \frac{P(B_2 \cap B_1')}{P(B_1')} = \frac{0.1}{0.5} = 0.2$ | M1 | Use of conditional probability
 | A1 (2) | 

## Part (d)
For independence: $P(B_1 \cap B_2) = P(B_1) \times P(B_2)$ | M1 | 
$P(B_1 \cap B_2) = 0.2$; $P(B_1) \times P(B_2) = 0.15$ | 0.2; 0.15 | A1
LHS $\neq$ RHS $\Rightarrow$ events not independent | | A1 (3)
NB: Accept alternate correct solutions | | 

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A contractor bids for two building projects. He estimates that the probability of winning the first project is 0.5, the probability of winning the second is 0.3 and the probability of winning both projects is 0.2.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that he does not win either project. [3]
\item Find the probability that he wins exactly one project. [2]
\item Given that he does not win the first project, find the probability that he wins the second. [2]
\item By calculation, determine whether or not winning the first contract and winning the second contract are independent events. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2002 Q4 [10]}}