| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Standard +0.3 This is a straightforward probability distribution question requiring systematic enumeration of outcomes (one biased coin, three fair coins), using sum of probabilities equals 1, and standard expectation calculation. Part (c) uses binomial distribution directly, and part (d) applies geometric distribution. All techniques are routine S1 content with no novel insight required, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| \(x\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 3 } { 80 }\) | \(a\) | \(b\) | \(c\) | \(\frac { 7 } { 80 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = P(1\ head) = 0.7\times(0.5)^3 + 0.3\times(0.5)^3\times 3 = \frac{1}{5}\) | B1 | Clear statement of unevaluated correct calculation \(= \frac{1}{5}\). AG |
| \(b = 0.7\times0.5^3\times3 + 0.3\times0.5^3\times3 = \frac{3}{8}\) | M1 | Clear statement of unevaluated calculation for either \(b\) or \(c\) |
| \(c = 0.7\times0.5^3\times3 + 0.3\times0.5^3 = \frac{3}{10}\) | A1 | For either \(b\) or \(c\) correct |
| \(\left[or\ c = \frac{27}{40} - b\right]\) | B1 FT | their \(b\) + their \(c = \frac{27}{40}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = \frac{3\times0 + 16\times1 + 30\times2 + 24\times3 + 7\times4}{80} = \frac{176}{80}\) or \(2.2\) | B1 FT | Correct or accept unsimplified calculation using their values for \(b\) and \(c\) seen (sum of probabilities \(= 1\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(0,1,2)] = ^{10}C_0\ 0.2^0\ 0.8^{10} + ^{10}C_1\ 0.2^1\ 0.8^9 + ^{10}C_2\ 0.2^2\ 0.8^8\) | M1 | One term \(^{10}C_x\ p^x(1-p)^{10-x}\), for \(0 < x < 10\), \(0 < p < 1\) |
| \(0.107374 + 0.268435 + 0.301989\) | A1 | Correct expression, accept unsimplified leading to final answer |
| \(0.678\) | B1 | \(0.677 < p \leqslant 0.678\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.8^6\times0.2 + 0.8^7\times0.2 = 0.0524288 + 0.041943\) | M1 | \(p^l\times(1-p) + p^m\times(1-p)\), \(l=6, 7\), \(m = l+1\), \(0 < p < 1\) |
| \(0.0944\) | A1 | \(0.09437 \leqslant p \leqslant 0.0944\) |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = P(1\ head) = 0.7\times(0.5)^3 + 0.3\times(0.5)^3\times 3 = \frac{1}{5}$ | B1 | Clear statement of unevaluated correct calculation $= \frac{1}{5}$. AG |
| $b = 0.7\times0.5^3\times3 + 0.3\times0.5^3\times3 = \frac{3}{8}$ | M1 | Clear statement of unevaluated calculation for either $b$ or $c$ |
| $c = 0.7\times0.5^3\times3 + 0.3\times0.5^3 = \frac{3}{10}$ | A1 | For either $b$ or $c$ correct |
| $\left[or\ c = \frac{27}{40} - b\right]$ | B1 FT | their $b$ + their $c = \frac{27}{40}$ |
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = \frac{3\times0 + 16\times1 + 30\times2 + 24\times3 + 7\times4}{80} = \frac{176}{80}$ or $2.2$ | B1 FT | Correct or accept unsimplified calculation using their values for $b$ and $c$ seen (sum of probabilities $= 1$) |
---
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(0,1,2)] = ^{10}C_0\ 0.2^0\ 0.8^{10} + ^{10}C_1\ 0.2^1\ 0.8^9 + ^{10}C_2\ 0.2^2\ 0.8^8$ | M1 | One term $^{10}C_x\ p^x(1-p)^{10-x}$, for $0 < x < 10$, $0 < p < 1$ |
| $0.107374 + 0.268435 + 0.301989$ | A1 | Correct expression, accept unsimplified leading to final answer |
| $0.678$ | B1 | $0.677 < p \leqslant 0.678$ |
---
## Question 4(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.8^6\times0.2 + 0.8^7\times0.2 = 0.0524288 + 0.041943$ | M1 | $p^l\times(1-p) + p^m\times(1-p)$, $l=6, 7$, $m = l+1$, $0 < p < 1$ |
| $0.0944$ | A1 | $0.09437 \leqslant p \leqslant 0.0944$ |4 Jacob has four coins. One of the coins is biased such that when it is thrown the probability of obtaining a head is $\frac { 7 } { 10 }$. The other three coins are fair. Jacob throws all four coins once. The number of heads that he obtains is denoted by the random variable $X$. The probability distribution table for $X$ is as follows.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 3 } { 80 }$ & $a$ & $b$ & $c$ & $\frac { 7 } { 80 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $a = \frac { 1 } { 5 }$ and find the values of $b$ and $c$.
\item Find $\mathrm { E } ( X )$.\\
Jacob throws all four coins together 10 times.
\item Find the probability that he obtains exactly one head on fewer than 3 occasions.
\item Find the probability that Jacob obtains exactly one head for the first time on the 7th or 8th time that he throws the 4 coins.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q4 [10]}}