CAIE S1 2023 March — Question 3 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2023
SessionMarch
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeGeometric distribution (first success)
DifficultyModerate -0.8 This is a straightforward binomial distribution question with standard probability calculations. Part (a) requires P(X>17) using binomial tables or calculator (routine). Parts (b) and (c) are geometric/negative binomial applications with simple probability multiplication. All parts are direct applications of formulas with no problem-solving insight required, making this easier than average.
Spec2.04b Binomial distribution: as model B(n,p)5.02f Geometric distribution: conditions
80\% of the residents of Kinwawa are in favour of a leisure centre being built in the town. 20 residents of Kinwawa are chosen at random and asked, in turn, whether they are in favour of the leisure centre.
  1. Find the probability that more than 17 of these residents are in favour of the leisure centre. [3]
  2. Find the probability that the 5th person asked is the first person who is not in favour of the leisure centre. [1]
  3. Find the probability that the 7th person asked is the second person who is not in favour of the leisure centre. [2]
Question 3:
AnswerMarks
3(a)Method 1 for Question 3(a)
[P(X > 17) = P(18, 19, 20) =]
20C ( 0.8 )18( 0.2 )2 +20C ( 0.8 )19( 0.2 )1
18 19
+ 20C ( 0.8 )20
20
AnswerMarks Guidance
= 0.13691 + 0.05765 + 0.01153M1 One term 20C ( p )x( 1− p )20−x , 0 p1,0 x20.
x
AnswerMarks
A1Correct expression, accept unsimplified, no terms omitted
leading to final answer.
AnswerMarks Guidance
0.206B1 Mark the final answer at the most accurate value 0.206 ⩽ p ⩽
0.2061 .
Method 2 for Question 3(a)
[P(X > 17) = 1 – P(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) = ]
1 – (20C ( 0.8 )0( 0.2 )20 +20C ( 0.8 )1( 0.2 )19
0 1
+ 20C ( 0.8 )2( 0.2 )18 ++ 20C ( 0.8 )16( 0.2 )4
2 16
+20C ( 0.8 )17( 0.2 )3 )
17
(1.04810−14 +8.38910−13
= 1 –
AnswerMarks Guidance
+3.1881011++0.2182+0.2054)M1 One term 20C ( p )x( 1− p )20−x , 0 p1, 0 x20.
x
AnswerMarks
A1Correct expression, accept unsimplified, no terms omitted
leading to final answer. If answer correct, condone omission of
any 15 of the 16 middle terms.
AnswerMarks Guidance
0.206B1 Mark the final answer at the most accurate value 0.206 ⩽ p ⩽
0.2061 .
Condone omission of brackets.
3
AnswerMarks
3(b)256
( )4( )=
0.8 0.2 0.08192,
 
AnswerMarks Guidance
3125B1 8192
Accept OE.
100000
1
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
3(c)( )5( )26
0.8 0.2M1 ( )5( )2k ( )5( )k0.2,
0.8 0.2 or 0.8 0.2
2 ⩽ k ⩽ 7.
8144
=0.0786,
AnswerMarks Guidance
78125A1 786432
0.0786 ⩽ p < 0.07865, .
10000000
If A0 awarded, SC B1 for correct answer WWW.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | Method 1 for Question 3(a)
[P(X > 17) = P(18, 19, 20) =]
20C ( 0.8 )18( 0.2 )2 +20C ( 0.8 )19( 0.2 )1
18 19
+ 20C ( 0.8 )20
20
= 0.13691 + 0.05765 + 0.01153 | M1 | One term 20C ( p )x( 1− p )20−x , 0 p1,0 x20.
x
A1 | Correct expression, accept unsimplified, no terms omitted
leading to final answer.
0.206 | B1 | Mark the final answer at the most accurate value 0.206 ⩽ p ⩽
0.2061 .
Method 2 for Question 3(a)
[P(X > 17) = 1 – P(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) = ]
1 – (20C ( 0.8 )0( 0.2 )20 +20C ( 0.8 )1( 0.2 )19
0 1
+ 20C ( 0.8 )2( 0.2 )18 ++ 20C ( 0.8 )16( 0.2 )4
2 16
+20C ( 0.8 )17( 0.2 )3 )
17
(1.04810−14 +8.38910−13
= 1 –
+3.1881011++0.2182+0.2054) | M1 | One term 20C ( p )x( 1− p )20−x , 0 p1, 0 x20.
x
A1 | Correct expression, accept unsimplified, no terms omitted
leading to final answer. If answer correct, condone omission of
any 15 of the 16 middle terms.
0.206 | B1 | Mark the final answer at the most accurate value 0.206 ⩽ p ⩽
0.2061 .
Condone omission of brackets.
3
--- 3(b) ---
3(b) | 256
( )4( )=
0.8 0.2 0.08192,
 
3125 | B1 | 8192
Accept OE.
100000
1
Question | Answer | Marks | Guidance
--- 3(c) ---
3(c) | ( )5( )26
0.8 0.2 | M1 | ( )5( )2k ( )5( )k0.2,
0.8 0.2 or 0.8 0.2
2 ⩽ k ⩽ 7.
8144
=0.0786,
78125 | A1 | 786432
0.0786 ⩽ p < 0.07865, .
10000000
If A0 awarded, SC B1 for correct answer WWW.
2
Question | Answer | Marks | Guidance
80\% of the residents of Kinwawa are in favour of a leisure centre being built in the town.

20 residents of Kinwawa are chosen at random and asked, in turn, whether they are in favour of the leisure centre.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that more than 17 of these residents are in favour of the leisure centre. [3]

\item Find the probability that the 5th person asked is the first person who is not in favour of the leisure centre. [1]

\item Find the probability that the 7th person asked is the second person who is not in favour of the leisure centre. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2023 Q3 [6]}}
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