| Date | May 2009 | Marks available | 2 | Reference code | 09M.1.sl.TZ2.13 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 13 | Adapted from | N/A |
Question
A weighted die has 2 red faces, 3 green faces and 1 black face. When the die is thrown, the black face is three times as likely to appear on top as one of the other five faces. The other five faces have equal probability of appearing on top.
The following table gives the probabilities.
Find the value of
(i) m;
(ii) n.
The die is thrown once.
Given that the face on top is not red, find the probability that it is black.
The die is now thrown twice.
Calculate the probability that black appears on top both times.
Markscheme
(i) m = 1 (A1)
(ii) n = 3 (A1) (C2)
Note: Award (A0)(A1)(ft) for \(m = \frac{1}{8}, n = \frac{3}{8}\).
Award (A0)(A1)(ft) for m = 3, n = 1.
[2 marks]
\({\rm{P}}(B/R') = \frac{{\frac{3}{8}}}{{\frac{6}{8}}} = \frac{3}{6}\left( {\frac{1}{2},50\% ,0.5} \right)\) (M1)(A1)(ft) (C2)
Note: Award (M1) for correctly substituted conditional probability formula or for 6 seen as part of denominator.
[2 marks]
\({\rm{P}}(B,B) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{{64}}(0.141)\) (M1)(A1)(ft) (C2)
Note: Award (M1) for product of two correct fractions, decimals or percentages.
(ft) from their answer to part (a) (ii).
[2 marks]
Examiners report
The answers 1/8 and 3/8 were provided by many rather than 1 and 3. The conditional probability question was correctly answered more often when the formula was used. A common incorrect answer to part (c) was 3/8 × 2/7.
The answers 1/8 and 3/8 were provided by many rather than 1 and 3. The conditional probability question was correctly answered more often when the formula was used. A common incorrect answer to part (c) was 3/8 × 2/7.
The answers 1/8 and 3/8 were provided by many rather than 1 and 3. The conditional probability question was correctly answered more often when the formula was used. A common incorrect answer to part (c) was 3/8 × 2/7.
