CAIE S1 2023 March — Question 2 7 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2023
SessionMarch
Marks7
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Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeMultiple unknowns from expectation and variance
DifficultyModerate -0.3 This is a straightforward discrete probability distribution question requiring systematic enumeration of outcomes (one biased coin, three fair coins), basic probability calculations, and standard variance formula application. While it involves multiple steps, each step uses routine S1 techniques with no novel insight required, making it slightly easier than average A-level.
Spec2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)
Alisha has four coins. One of these coins is biased so that the probability of obtaining a head is 0.6. The other three coins are fair. Alisha throws the four coins at the same time. The random variable \(X\) denotes the number of heads obtained.
  1. Show that the probability of obtaining exactly one head is 0.225. [3]
  2. Complete the following probability distribution table for \(X\). [2]
    \(x\)01234
    P(\(X = x\))0.050.2250.075
  3. Given that E(\(X\)) = 2.1, find the value of Var(\(X\)). [2]
Question 2:
AnswerMarks Guidance
2(a)0.6(0.5)3 + 0.4 ( 0.5 )33 B1
or b + 0.4 ( 0.5 )3 ( 3 or 3C ) , 0 < a, b < 1 seen.
1
AnswerMarks
M10.6(0.5)3 + 0.4 ( 0.5 )3d seen, d = 1, 3.
Condone 0.075 + 0.05 × d, d = 1, 3.
AnswerMarks Guidance
= 0.225A1 AG full supporting working required.
Scenarios identified and linked to calculations.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
2(b)x 0 1 2 3 4
P(X = x) 0.05 0.225 0.375 0.275 0.075B1 3
Either [P(2) =] 0.375,
8
11
or [P(3) =] 0.275, seen.
40
Condone not in table if identified.
AnswerMarks
B1 FTBoth values in table.
FT P(2) + P(3) = 0.650 .
2
AnswerMarks Guidance
2(c)Var(X)
= [12]0.225+22their 0.375+32their 0.275+420.075−2.12M1 Appropriate variance formula from their probability
distribution table with at least 4 terms, 0 < their P(x) < 1.
Condone 4.41 for 2.12 .
Condone mean clearly recalculated inaccurately.
Or
0.225+4their 0.375+9their 0.275+160.075−2.12
Condone 2.12 for 4.41.
5.4−2.12=0.99  0 
AnswerMarks Guidance
 A1 If M0 awarded
SC B1 for 0.99[0] WWW.
2
AnswerMarks Guidance
x0 1
P(X = x)0.05 0.225
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | 0.6(0.5)3 + 0.4 ( 0.5 )33 | B1 | Either 0.6(0.5)3 + a
or b + 0.4 ( 0.5 )3 ( 3 or 3C ) , 0 < a, b < 1 seen.
1
M1 | 0.6(0.5)3 + 0.4 ( 0.5 )3d seen, d = 1, 3.
Condone 0.075 + 0.05 × d, d = 1, 3.
= 0.225 | A1 | AG full supporting working required.
Scenarios identified and linked to calculations.
3
Question | Answer | Marks | Guidance
--- 2(b) ---
2(b) | x 0 1 2 3 4
P(X = x) 0.05 0.225 0.375 0.275 0.075 | B1 | 3
Either [P(2) =] 0.375,
8
11
or [P(3) =] 0.275, seen.
40
Condone not in table if identified.
B1 FT | Both values in table.
FT P(2) + P(3) = 0.650 .
2
--- 2(c) ---
2(c) | Var(X)
= [12]0.225+22their 0.375+32their 0.275+420.075−2.12 | M1 | Appropriate variance formula from their probability
distribution table with at least 4 terms, 0 < their P(x) < 1.
Condone 4.41 for 2.12 .
Condone mean clearly recalculated inaccurately.
Or
0.225+4their 0.375+9their 0.275+160.075−2.12
Condone 2.12 for 4.41.
5.4−2.12=0.99  0 
  | A1 | If M0 awarded
SC B1 for 0.99[0] WWW.
2
x | 0 | 1 | 2 | 3 | 4
P(X = x) | 0.05 | 0.225 | 0.375 | 0.275 | 0.075
Question | Answer | Marks | Guidance
Alisha has four coins. One of these coins is biased so that the probability of obtaining a head is 0.6. The other three coins are fair. Alisha throws the four coins at the same time. The random variable $X$ denotes the number of heads obtained.

\begin{enumerate}[label=(\alph*)]
\item Show that the probability of obtaining exactly one head is 0.225. [3]

\item Complete the following probability distribution table for $X$. [2]

\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
P($X = x$) & 0.05 & 0.225 & & & 0.075 \\
\hline
\end{tabular}

\item Given that E($X$) = 2.1, find the value of Var($X$). [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2023 Q2 [7]}}
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