Date | May 2018 | Marks available | 2 | Reference code | 18M.2.sl.TZ1.5 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Contestants in a TV gameshow try to get through three walls by passing through doors without falling into a trap. Contestants choose doors at random.
If they avoid a trap they progress to the next wall.
If a contestant falls into a trap they exit the game before the next contestant plays.
Contestants are not allowed to watch each other attempt the game.
The first wall has four doors with a trap behind one door.
Ayako is a contestant.
Natsuko is the second contestant.
The second wall has five doors with a trap behind two of the doors.
The third wall has six doors with a trap behind three of the doors.
The following diagram shows the branches of a probability tree diagram for a contestant in the game.
Write down the probability that Ayako avoids the trap in this wall.
Find the probability that only one of Ayako and Natsuko falls into a trap while attempting to pass through a door in the first wall.
Copy the probability tree diagram and write down the relevant probabilities along the branches.
A contestant is chosen at random. Find the probability that this contestant fell into a trap while attempting to pass through a door in the second wall.
A contestant is chosen at random. Find the probability that this contestant fell into a trap.
120 contestants attempted this game.
Find the expected number of contestants who fell into a trap while attempting to pass through a door in the third wall.
Markscheme
\(\frac{3}{4}\) (0.75, 75%) (A1)
[1 mark]
\(\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}\) OR \(2 \times \frac{3}{4} \times \frac{1}{4}\) (M1)(M1)
Note: Award (M1) for their product \(\frac{1}{4} \times \frac{3}{4}\) seen, and (M1) for adding their two products or multiplying their product by 2.
\( = \frac{3}{8}\,\,\,\,\left( {\frac{6}{{16}},\,\,0.375,\,\,37.5{\text{% }}} \right)\) (A1)(ft) (G3)
Note: Follow through from part (a), but only if the sum of their two fractions is 1.
[3 marks]
(A1)(ft)(A1)(A1)
Note: Award (A1) for each correct pair of branches. Follow through from part (a).
[3 marks]
\(\frac{3}{4} \times \frac{2}{5}\) (M1)
Note: Award (M1) for correct probabilities multiplied together.
\( = \frac{3}{{10}}\,\,\,\left( {0.3,\,\,30{\text{% }}} \right)\) (A1)(ft) (G2)
Note: Follow through from their tree diagram or part (a).
[2 marks]
\(1 - \frac{3}{4} \times \frac{2}{5} \times \frac{3}{6}\) OR \(\frac{1}{4} + \frac{3}{4} \times \frac{2}{5} + \frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\) (M1)(M1)
Note: Award (M1) for \(\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\) and (M1) for subtracting their correct probability from 1, or adding to their \(\frac{1}{4} + \frac{3}{4} \times \frac{2}{5}\).
\( = \frac{{93}}{{120}}\,\,\,\,\left( {\frac{{31}}{{40}},\,\,0.775,\,\,77.5{\text{% }}} \right)\) (A1)(ft) (G2)
Note: Follow through from their tree diagram.
[3 marks]
\(\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6} \times 120\) (M1)(M1)
Note: Award (M1) for \(\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\,\,\,\,\left( {\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\,\,{\text{OR}}\,\,\frac{{27}}{{120}}\,\,{\text{OR}}\,\,\frac{9}{{40}}} \right)\) and (M1) for multiplying by 120.
= 27 (A1)(ft) (G3)
Note: Follow through from their tree diagram or their \(\frac{3}{4} \times \frac{3}{5} \times \frac{3}{6}\) from their calculation in part (d)(ii).
[3 marks]