S2 January 2009 Q5
5. A factory produces components of which 1% are defective. The components are packed in boxes of 10. A box is selected at random.
(a) Find the probability that the box contains exactly one defective component. (2)
(b) Find the probability that there are at least 2 defective components in the box. (3)
(c) Using a suitable approximation, find the probability that a batch of 250 components contains between 1 and 4 (inclusive) defective components. (4)
| Scheme | Marks |
|---|---|
| \(X\) represents the number of defective components. | |
| \(\mathrm{P}(X = 1) = (0.99)^9(0.01) \times 10 = 0.0914\) | M1A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \geqslant 2) = 1 - \mathrm{P}(X \leqslant 1)\) | M1 |
| \(= 1 - (p)^{10} - (a)\) | A1ft |
| \(= 0.0043\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(2.5)\) | B1B1 |
| \(\mathrm{P}(1 \leqslant X \leqslant 4) = \mathrm{P}(X \leqslant 4) - \mathrm{P}(X = 0)\) \(= 0.8912 - 0.0821\) | M1 |
| \(= 0.809\) | A1 |
| (4) | |
| (9 marks) |
Notes
Normal distribution used. B1for mean only
Special case for parts a and b
If they use 0.1 do not treat as misread as it makes it easier.
(a) M1 A0 if they have 0.3874
(b) M1 A1ft A0 they will get 0.2639
(c) Could get B1 B0 M1 A0
For any other values of \(p\) which are in the table do not use misread. Check using the tables. They could get (a) M1 A0 (b) M1 A1ft A0 (c) B1 B0 M1 A0