| Date | November 2013 | Marks available | 2 | Reference code | 13N.2.hl.TZ0.5 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
At the start of each week, Eric and Marina pick a night at random on which they will watch a movie.
If they choose a Saturday night, the probability that they watch a French movie is \(\frac{7}{9}\) and if they choose any other night the probability that they watch a French movie is \(\frac{4}{9}\).
Find the probability that they watch a French movie.
Given that last week they watched a French movie, find the probability that it was on a Saturday night.
Markscheme
\({\text{P}}(F) = \left( {\frac{1}{7} \times \frac{7}{9}} \right) + \left( {\frac{6}{7} \times \frac{4}{9}} \right)\) (M1)(A1)
Note: Award M1 for the sum of two products.
\( = \frac{{31}}{{63}}{\text{ }}( = 0.4920 \ldots )\) A1
[3 marks]
Use of \({\text{P}}(S|F) = \frac{{{\text{P}}(S \cap F)}}{{{\text{P}}(F)}}\) to obtain \({\text{P}}(S|F) = \frac{{\frac{1}{7} \times \frac{7}{9}}}{{\frac{{31}}{{63}}}}\). M1
Note: Award M1 only if the numerator results from the product of two probabilities.
\( = \frac{7}{{31}}{\text{ }}( = 0.2258 \ldots )\) A1
[2 marks]
Examiners report
Both parts were very well done. In part (a), most candidates successfully used a tree diagram.
Both parts were very well done. In part (b), most candidates correctly used conditional probability considerations.
