S2 January 2010 Q3
3. A robot is programmed to build cars on a production line. The robot breaks down at random at a rate of once every 20 hours.
Find the probability that, in an 8 hour period,
In a particular 8 hour period, the robot broke down twice.
| Scheme | Marks |
|---|---|
| \(Y \sim \mathrm{Po}(0.25)\) | B1 |
| \(\mathrm{P}(Y = 0) = \mathrm{e}^{-0.25}\) | M1 |
| \(= 0.7788\) | A1 |
| (3) |
Notes
B1 for seeing or using Po(0.25)
M1 for finding \(\mathrm{P}(Y = 0)\) either by \(\mathrm{e}^{-a}\), where \(a\) is positive (\(a\) needn’t equal their \(\lambda\)) or using tables if their value of \(\lambda\) is in them
Beware common Binomial error using, \(p = 0.05\) gives 0.7738 but scores B0 M0 A0
A1 awrt 0.779
SC Use of Binomial. Mark parts a and b as scheme. They could get (a) B0,M0,A0 (b) B0 M1 A0
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(0.4)\) | B1 |
| \(\mathrm{P}(\text{Robot will break down}) = 1 - \mathrm{P}(X = 0)\) \(= 1 - e^{-0.4}\) | M1 |
| \(= 1 - 0.067032\) \(= 0.3297\) | A1 |
| (3) |
Notes
B1 for stating or a clear use of Po(0.4) in part (b) or (c)
M1 for writing or finding \(1 - \mathrm{P}(X = 0)\)
A1 awrt 0.33
SC Use of Binomial. Mark parts a and b as scheme. They could get (a) B0,M0,A0 (b) B0 M1 A0
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X = 2) = \dfrac{e^{-0.4}(0.4)^2}{2}\) | M1 |
| \(= 0.0536\) | A1 |
| (2) |
Notes
M1 for finding \(\mathrm{P}(X = 2)\) e.g \(\dfrac{\mathrm{e}^{-\lambda}\lambda^2}{2!}\) with their value of \(\lambda\) in
or if their \(\lambda\) is in the table for writing \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1)\)
A1 awrt 0.0536
SC Use of Binomial. In part c allow M1 for \({}^{n}C_2\,(p)^2(1 - p)^{n-2}\) with “their n” and “their \(p\)”. They could get (c) M1,A0
DO NOT GIVE for \(p(x \leqslant 2) - p(x \leqslant 1)\)
| Scheme | Marks |
|---|---|
| 0.3297 or answer to part (b) as Poisson events are independent | B1ft B1 dep |
| (2) | |
| (10 marks) |
Notes
1st B1 their answer to part(b) correct to 2 sf or awrt 0.33
2nd B1 need the word independent. This is dependent on them gaining the first B1
SC Use of Binomial. In (d) they can get the first B1 only. They could get (d) B1B0