S2 January 2008 Q2
2. The probability of a bolt being faulty is 0.3. Find the probability that in a random sample of 20 bolts there are
These bolts are sold in bags of 20. John buys 10 bags.
| Scheme | Marks |
|---|---|
| Let \(X\) be the random variable the number of faulty bolts | |
| \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1) = 0.0355 - 0.0076\) or \((0.3)^2(0.7)^{18}\dfrac{20!}{18!\,2!}\) | M1 |
| \(= 0.0279\) \(= 0.0278\) | A1 |
| (2) |
Notes
M1 Either attempting to use \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1)\)
or attempt to use binomial and find \(\mathrm{p}(X = 2)\). Must have \((p)^2(1 - p)^{18}\dfrac{20!}{18!\,2!}\), with a value of \(p\)
A1 awrt 0.0278 or 0.0279.
| Scheme | Marks |
|---|---|
| \(1 - \mathrm{P}(X \leqslant 3) = 1 - 0.1071\) | M1 |
| \(= 0.8929\) or \(1 - (0.3)^3(0.7)^{17}\dfrac{20!}{17!\,3!} - (0.3)^2(0.7)^{18}\dfrac{20!}{18!\,2!} - (0.3)(0.7)^{19}\dfrac{20!}{19!\,1!} - (0.7)^{20}\) | A1 |
| (2) |
Notes
M1 Attempting to find \(1 - \mathrm{P}(X \leqslant 3)\)
A1 awrt 0.893
| Scheme | Marks |
|---|---|
| \(\dfrac{10!}{4!\,6!}(0.8929)^6(0.1071)^4 = 0.0140.\) | M1A1ftA1 |
| (3) | |
| (7 marks) |
Notes
M1 for \(k\,(p)^6(1 - p)^4\). They may use any value for \(p\) and \(k\) can be any number or \({}^n\mathrm{C}_6\,p^6(1 - p)^{n-6}\)
A1ft \(\dfrac{10!}{4!\,6!}(\textit{their part b})^6(1 - \textit{their part b})^4\) may write \({}^{10}\mathrm{C}_6\) or \({}^{10}\mathrm{C}_4\)
A1 awrt 0.014