| Date | November 2011 | Marks available | 2 | Reference code | 11N.2.hl.TZ0.5 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
The probability that the 08:00 train will be delayed on a work day (Monday to Friday) is \(\frac{1}{{10}}\). Assuming that delays occur independently,
find the probability that the 08:00 train is delayed exactly twice during any period of five work days;
find the minimum number of work days for which the probability of the 08:00 train being delayed at least once exceeds 90 %.
Markscheme
\(X \sim {\text{B(5, 0.1)}}\) (M1)
\({\text{P}}(X = 2) = 0.0729\) A1
[2 marks]
\({\text{P}}(X \geqslant 1) = 1 - {\text{P}}(X = 0)\) (M1)
\(0.9 < 1 - {\left( {\frac{9}{{10}}} \right)^n}\) (M1)
\(n > \frac{{\ln 0.1}}{{\ln 0.9}}\)
n = 22 days A1
[3 marks]
Examiners report
This question was generally answered successfully. Many candidates used the tabular feature of their GDC for (b) thereby avoiding potential errors in the algebraic manipulation of logs and inequalities.
This question was generally answered successfully. Many candidates used the tabular feature of their GDC for (b) thereby avoiding potential errors in the algebraic manipulation of logs and inequalities.
