S2 June 2014 Q4
4. A cadet fires shots at a target at distances ranging from 25 m to 90 m. The probability of hitting the target with a single shot is \(p\). When firing from a distance \(d\) m, \(p = \dfrac{3}{200}(90 - d)\).
Each shot is fired independently.
The cadet fires 10 shots from a distance of 40 m.
The cadet fires 20 shots from a distance of \(x\) m.
The cadet fires 100 shots from 25 m.
| Scheme | Marks |
|---|---|
| \(X\) is the random variable the Number of successes, \(X \sim \mathrm{B}(10, 0.75)\) | B1 |
| (i) \(\mathrm{P}(X = 6) = (0.75)^6(0.25)^4\,{}^{10}C_6\) or \(\mathrm{P}(X \leqslant 6) - \mathrm{P}(X \leqslant 5)\) | M1 |
| \(= 0.145998\) awrt 0.146 | A1 |
| (ii) Using \(X \sim \mathrm{B}(10, 0.75)\) \(\mathrm{P}(X \geqslant 8) = \mathrm{P}(X = 8) + \mathrm{P}(X = 9) + \mathrm{P}(X = 10)\) | M1 |
| \(= (0.75)^8(0.25)^2\,{}^{10}C_8 + (0.75)^9(0.25)^1\,{}^{10}C_9 + (0.75)^{10}\) | |
| \(= 0.52559\) awrt 0.526 | A1 |
| Or Using \(Y \sim \mathrm{B}(10, 0.25)\) and \(\mathrm{P}(Y \leqslant 2) = 0.5256\) | |
| (5) |
Notes
B1 writing or using \(p = 0.75\) or \(p = 0.25\) anywhere in (a)(i) or (a)(ii)
(i) M1 writing or using \((p)^6(1 - p)^4\,{}^{10}C_6\) or writing for \(p = 0.75\), \(\mathrm{P}(X \leqslant 6) - (X \leqslant 5)\) or for \(p = 0.25\), \(\mathrm{P}(X \leqslant 4) - \mathrm{P}(X \leqslant 3)\) or correct answer.
(ii) M1 writing B(10, 0.75) and writing or using \(\mathrm{P}(X = 8) + \mathrm{P}(X = 9) + \mathrm{P}(X = 10)\) oe or writing B(10, 0.25) and writing or using \(\mathrm{P}(Y \leqslant 2)\).
Using correct Binomial must be shown by \((0.75)^n(0.25)^{10 - n}\) or a correct answer.
| Scheme | Marks |
|---|---|
| \(1 - \mathrm{P}(0) = 0.8\) or \(\mathrm{P}(0) = 0.2\) | M1 |
| \((1 - p)^{20} = 0.2\) | |
| \(1 - p = 0.9227\) | |
| \(p = 0.0773\) | A1 |
| \(\dfrac{3}{200}(90 - x) = 0.0773\) | M1 |
| \(x = 84.84\) | |
| \(x = 85\) | A1cao |
| (4) |
Notes
M1 for writing or using 1 – P(0) = 0.8 or P(0) = 0.2 or \((1 - p)^{20} = 0.2\). Allow any inequality sign.
A1 awrt 0.0773 or awrt 0.923.
M1 subst in \(\dfrac{3}{200}(90 - x)\) for \(p\) NB this may be substituted in earlier for \(p\).
Allow for \(\dfrac{3}{200}(90 - x) = k\) where \(0 \lt k \lt 1\) \(k \ne 0.8\) or 0.2 Allow any inequality sign
A1 condone \(x \geqslant 85\). Do not allow \(x \leqslant 85\).
| Scheme | Marks |
|---|---|
| \(X\) – successes \(\sim \mathrm{B}(100, 0.975)\) \(Y\) – not successes \(\sim \mathrm{B}(100, 0.025)\) | B1 |
| \(Y \sim \mathrm{Po}(2.5)\) | M1A1 |
| \(\mathrm{P}(Y \leqslant 5) = 0.958\) | M1A1 |
| (5) | |
| (14 marks) |
Notes
B1 writing or using 0.975 or 0.025, may be implied by Po(2.5)
M1 using Po approximation
A1 Po(2.5)
M1 writing or using \(\mathrm{P}(Y \leqslant 5)\)
A1 awrt 0.958
SC use of normal approximation can get B1 M0A0M1A0
B1 writing or using 0.975 or 0.025 implied by normal with mean 97.5 or answer of 0.973
M1 for awrt 0.973