| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Given conditional, find joint or marginal |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question with standard calculations. Part (a) requires finding P(B) using complement rule (shown answer guides students), part (b) applies the conditional probability formula P(A|B) = P(A∩B)/P(B), and part (c) involves listing outcomes systematically. All techniques are routine for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(BH_1\;BT_2) = \frac{1}{4}\times\frac{3}{4} = \frac{3}{16}\); \(P(BT_1\;BH_2) = \frac{3}{4}\times\frac{1}{4} = \frac{3}{16}\); \(P(BH_1\;BH_2) = \frac{1}{4}\times\frac{1}{4} = \frac{1}{16}\) | M1 | All 3 different calculations seen unsimplified |
| \(\frac{3}{16}+\frac{3}{16}+\frac{1}{16} = \frac{7}{16}\) | A1 | Clear identification of all scenarios, linked probabilities and sum. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(H\ BH_1\ BT_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32}\) | M1 | All 6 different calculations seen unsimplified |
| \(P(T\ BH_1\ BT_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32}\) | ||
| \(P(H\ BT_1\ BH_2) = \frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}\) | ||
| \(P(T\ BT_1\ BH_2) = \frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}\) | ||
| \(P(H\ BH_1\ BH_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}\) | ||
| \(P(T\ BH_1\ BH_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}\) | ||
| \(P(B) = \frac{1+3+3+1+3+3}{32} = \frac{14}{32} = \frac{7}{16}\) | A1 | Clear identification of all scenarios, linked probabilities and sum. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - P(BT_1\ BT_2) = 1 - \left(\frac{3}{4}\right)^2\) | M1 | Calculation seen unsimplified and \(1-\) probability seen |
| \(= \frac{7}{16}\) | A1 | Clear identification of scenario used, linked probability and calculation. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1 - P(H\ BT_1\ BT_2) - P(T\ BT_1\ BT_2) = 1 - \left(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4}\right) - \left(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4}\right)\) | M1 | Both calculations seen unsimplified and \(1-2\) probabilities seen |
| \(= 1 - \frac{9}{32} - \frac{9}{32} = \frac{7}{16}\) | A1 | Clear identification of all scenarios used, linked probabilities and calculation. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(A\ | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{7}{16}} = \frac{\frac{1}{32}}{\frac{7}{16}}\) | M1 |
| \(= \frac{1}{14},\ 0.0714\) | A1 | Accept \(0.071428\ldots\) rounded to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(1H) = \frac{1}{2}\times\frac{3}{4}\times\frac{1}{4} + \frac{1}{2}\times\frac{1}{4}\times\frac{3}{4} + \frac{1}{2}\times\frac{3}{4}\times\frac{3}{4} = \frac{15}{32}\) | B1 | Table with correct \(X\) values and at least one probability. Condone any additional \(X\) values if probability stated as 0 |
| \(P(2H) = \frac{1}{2}\times\frac{1}{4}\times\frac{1}{4} + \frac{1}{2}\times\frac{3}{4}\times\frac{1}{4} + \frac{1}{2}\times\frac{1}{4}\times\frac{3}{4} = \frac{7}{32}\) | B1 | \(P(1)\) or \(P(2)\) correct, need not be in table, accept unsimplified |
| \(\begin{array}{c\ | cccc} X & 0 & 1 & 2 & 3 \\ \hline p(X) & \frac{9}{32} & \frac{15}{32} & \frac{7}{32} & \frac{1}{32} \end{array}\) | B1 |
| SC B1 for 4 probabilities \((0 < p < 1)\) sum to \(1 \pm 0.005\) with \(P(1)\) and \(P(2)\) incorrect |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(BH_1\;BT_2) = \frac{1}{4}\times\frac{3}{4} = \frac{3}{16}$; $P(BT_1\;BH_2) = \frac{3}{4}\times\frac{1}{4} = \frac{3}{16}$; $P(BH_1\;BH_2) = \frac{1}{4}\times\frac{1}{4} = \frac{1}{16}$ | M1 | All 3 different calculations seen unsimplified |
| $\frac{3}{16}+\frac{3}{16}+\frac{1}{16} = \frac{7}{16}$ | A1 | Clear identification of **all scenarios**, linked probabilities and sum. AG |
## Question 5(a):
**Method 2: Scenarios identified with all 3 coins**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(H\ BH_1\ BT_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32}$ | M1 | All 6 different calculations seen unsimplified |
| $P(T\ BH_1\ BT_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32}$ | | |
| $P(H\ BT_1\ BH_2) = \frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}$ | | |
| $P(T\ BT_1\ BH_2) = \frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}$ | | |
| $P(H\ BH_1\ BH_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}$ | | |
| $P(T\ BH_1\ BH_2) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}$ | | |
| $P(B) = \frac{1+3+3+1+3+3}{32} = \frac{14}{32} = \frac{7}{16}$ | A1 | Clear identification of **all scenarios**, linked probabilities and sum. AG |
**Method 3: $1 - P(BT_1\ BT_2)$ ignoring unbiased coin**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - P(BT_1\ BT_2) = 1 - \left(\frac{3}{4}\right)^2$ | M1 | Calculation seen unsimplified **and** $1-$ probability seen |
| $= \frac{7}{16}$ | A1 | Clear identification of scenario used, linked probability and calculation. AG |
**Method 4: $1 - P(BT_1\ BT_2)$ with all 3 coins**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1 - P(H\ BT_1\ BT_2) - P(T\ BT_1\ BT_2) = 1 - \left(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4}\right) - \left(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4}\right)$ | M1 | Both calculations seen unsimplified **and** $1-2$ probabilities seen |
| $= 1 - \frac{9}{32} - \frac{9}{32} = \frac{7}{16}$ | A1 | Clear identification of **all scenarios** used, linked probabilities and calculation. AG |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(A\|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{7}{16}} = \frac{\frac{1}{32}}{\frac{7}{16}}$ | M1 | *Their* identified $P(HHH)$ or correct as numerator **and** *their* identified $P(B)$ or correct as denominator. Either numerical expression acceptable |
| $= \frac{1}{14},\ 0.0714$ | A1 | Accept $0.071428\ldots$ rounded to at least 3SF |
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(1H) = \frac{1}{2}\times\frac{3}{4}\times\frac{1}{4} + \frac{1}{2}\times\frac{1}{4}\times\frac{3}{4} + \frac{1}{2}\times\frac{3}{4}\times\frac{3}{4} = \frac{15}{32}$ | B1 | Table with correct $X$ values and at least one probability. Condone any additional $X$ values if probability stated as 0 |
| $P(2H) = \frac{1}{2}\times\frac{1}{4}\times\frac{1}{4} + \frac{1}{2}\times\frac{3}{4}\times\frac{1}{4} + \frac{1}{2}\times\frac{1}{4}\times\frac{3}{4} = \frac{7}{32}$ | B1 | $P(1)$ or $P(2)$ correct, need not be in table, accept unsimplified |
| $\begin{array}{c\|cccc} X & 0 & 1 & 2 & 3 \\ \hline p(X) & \frac{9}{32} & \frac{15}{32} & \frac{7}{32} & \frac{1}{32} \end{array}$ | B1 | 4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3 SF |
| | | **SC B1** for 4 probabilities $(0 < p < 1)$ sum to $1 \pm 0.005$ with $P(1)$ **and** $P(2)$ incorrect |
---5 Eric has three coins. One of the coins is fair. The other two coins are each biased so that the probability of obtaining a head on any throw is $\frac { 1 } { 4 }$, independently of all other throws. Eric throws all three coins at the same time.
Events $A$ and $B$ are defined as follows.\\
$A$ : all three coins show the same result\\
$B$ : at least one of the biased coins shows a head
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( B ) = \frac { 7 } { 16 }$.
\item Find $\mathrm { P } ( A \mid B )$.\\
The random variable $X$ is the number of heads obtained when Eric throws the three coins.
\item Draw up the probability distribution table for $X$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q5 [7]}}