| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Most likely value (mode) |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question requiring standard techniques: (i) solving an inequality with logarithms, (ii) finding the mode using the given hint about values near the mean, and (iii) applying binomial probability twice. All parts are routine applications with no novel insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((0.75)^n < 0.06\) | M1* | Equation or inequality with 0.75\(^n\) and 0.06 or 0.94 seen |
| \(n > 9.78\) | M1 dep* | Attempt at solving by trial and error (can be implied) or using logarithms correctly |
| \(n = 10\) | A1 [3] | Correct answer |
| (ii) \(E(X) = 14 \times 0.75\) or \(10.5\) | M1 | Evaluating binomial probability for an integer value directly above or below their mean |
| Try \(P(10) = {}^{14}C_{10}(0.75)^{10}(0.25)^4 = 0.220\) | M1 | Evaluating the other binomial probability |
| \(P(11) = {}^{14}C_{11}(0.75)^{11}(0.25)^3 = 0.240\) (mode is) 11 | A1 [3] | Correct answer |
| OR | M1 | Evaluating binomial \(P(n)\) and \(P(n+1)\) |
| M1 | Evaluating binomial \(P(10)\), \(P(11)\) and \(P(12)\) | |
| A1 | Correct answer | |
| (iii) \(P(\geq 11) = {}^{14}C_{11}(0.75)^{12}(0.25)^2 + {}^{14}C_{13}(0.75)^{13}(0.25)^1 + (0.75)^{14}\) | M1 | A binomial term of the form \(^nC_rp^r(1-p)^{14-r}\) seen, \(n \neq 0\) or 14 |
| M1 | Summing binomial \(P(12, 13, 14)\) or \(P(11, 12, 13, 14)\) | |
| \(= 0.281\) | A1 | Correct 0.280 – 0.282 |
| \(P(3) = ^5C_3(0.2811)^3(0.7189)^2\) | M1 | A binomial term of the form \(^kC_rp^r(1-p)^2\) seen, any \(p\) |
| \(= 0.115\) | A1 [5] | Correct answer |
**(i)** $(0.75)^n < 0.06$ | M1* | Equation or inequality with 0.75$^n$ and 0.06 or 0.94 seen
$n > 9.78$ | M1 dep* | Attempt at solving by trial and error (can be implied) or using logarithms correctly
$n = 10$ | A1 [3] | Correct answer
**(ii)** $E(X) = 14 \times 0.75$ or $10.5$ | M1 | Evaluating binomial probability for an integer value directly above or below their mean
Try $P(10) = {}^{14}C_{10}(0.75)^{10}(0.25)^4 = 0.220$ | M1 | Evaluating the other binomial probability
$P(11) = {}^{14}C_{11}(0.75)^{11}(0.25)^3 = 0.240$ (mode is) 11 | A1 [3] | Correct answer
OR | M1 | Evaluating binomial $P(n)$ and $P(n+1)$
| M1 | Evaluating binomial $P(10)$, $P(11)$ and $P(12)$
| A1 | Correct answer
**(iii)** $P(\geq 11) = {}^{14}C_{11}(0.75)^{12}(0.25)^2 + {}^{14}C_{13}(0.75)^{13}(0.25)^1 + (0.75)^{14}$ | M1 | A binomial term of the form $^nC_rp^r(1-p)^{14-r}$ seen, $n \neq 0$ or 14
| M1 | Summing binomial $P(12, 13, 14)$ or $P(11, 12, 13, 14)$
$= 0.281$ | A1 | Correct 0.280 – 0.282
$P(3) = ^5C_3(0.2811)^3(0.7189)^2$ | M1 | A binomial term of the form $^kC_rp^r(1-p)^2$ seen, any $p$
$= 0.115$ | A1 [5] | Correct answer6 The probability that Sue completes a Sudoku puzzle correctly is 0.75 .\\
(i) Sue attempts $n$ Sudoku puzzles. Find the least value of $n$ for which the probability that she completes all $n$ puzzles correctly is less than 0.06 .
Sue attempts 14 Sudoku puzzles every month. The number that she completes successfully is denoted by $X$.\\
(ii) Find the value of $X$ that has the highest probability. You may assume that this value is one of the two values closest to the mean of $X$.\\
(iii) Find the probability that in exactly 3 of the next 5 months Sue completes more than 11 Sudoku puzzles correctly.
\hfill \mbox{\textit{CAIE S1 2011 Q6 [11]}}