Foundation June 2018 Paper 2 Q25
25 On three days, Ali throws darts at a target.
Here are his results.
| Number of throws | Number of hits | Number of misses | |
|---|---|---|---|
| Monday | 20 | 15 | 5 |
| Tuesday | 30 | 22 | 8 |
| Wednesday | 40 | 17 | 23 |
| Total | 90 | 54 | 36 |
(a) Work out two different estimates for the probability of Ali hitting the target. [2 marks]
(b) Which of your two answers is the better estimate for the probability of Ali hitting the target?
Give a reason for your answer. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Two different probabilities from \(\dfrac{15}{20}\) or 0.75 or 75% or \(\dfrac{22}{30}\) or 0.73… or 73.(…)% or \(\dfrac{17}{40}\) or 0.425 or 0.43 or 42.5% or 43% or \(\dfrac{54}{90}\) or 0.6 or 60% or \(\dfrac{37}{50}\) or 0.74 or 74% or \(\dfrac{32}{60}\) or 0.53… or 53.(…)% or \(\dfrac{39}{70}\) or 0.557… or 0.56 or 55.7…% or 56% | B2 | oe B1 for one correct probability |
Additional guidance
| Accept \(\dfrac{108}{180}\) as one of the probabilities | |
| Mark the answer line if it has two answers ignoring any incorrect probabilities in the working lines | |
| Ignore any incorrect cancelling or change of form (fraction, decimal or percentage) | |
| If the answer line only has one answer, check the working lines for a second answer for B2. Ignore any extra probabilities, unless incorrect, in which case award B1 max | |
| eg Working lines \(\dfrac{15}{20}\) Answer line \(\dfrac{54}{90}\) | B2 |
| eg Working lines \(\dfrac{15}{20}\), \(\dfrac{5}{15}\) Answer line \(\dfrac{54}{90}\) | B1 |
| If the answer line is blank, check the working lines for answers for B1 or B2. Ignore any extra probabilities, unless incorrect, in which case award B1 max | |
| eg Working lines \(\dfrac{15}{20}\), \(\dfrac{22}{30}\), \(\dfrac{54}{90}\) Answer line blank | B2 |
| eg Working lines \(\dfrac{15}{20}\), \(\dfrac{5}{15}\), \(\dfrac{54}{90}\) Answer line blank | B1 |
| Probabilities must not be given as ratios | |
| Do not accept the average of the given probabilities as answer |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 (ft their part (a)) | ||
| Their probability with the greater number of trials and valid reason eg More throws | B1ft | ft their two different probabilities from part (a) both probabilities must have a denominator based on throws |
| Alternative method 2 (independent of part (a)) | ||
| \(\dfrac{54}{90}\) and valid reason eg Total throws | B1 | oe |
Additional guidance
| Accept any unambiguous indication of their probability eg the day | |
| Using ratios | B0 |
| Ignore any non-contradictory statements | |
| 60% and It’s for all three days | B1 |
| \(\dfrac{54}{90}\) and It takes into account more throws | B1 |
| \(\dfrac{17}{40}\) (with \(\dfrac{22}{30}\) also in (a)) and Because he threw it more on Wednesday | B1ft |
| \(\dfrac{54}{90}\) and Shows the overall probability | B1 |
| \(\dfrac{54}{90}\) and Probability over total throws | B1 |
| \(\dfrac{54}{90}\) (with Wednesday probability in (a)) and It’s the average total days, not just Wednesdays | B1ft |
| Correct ft probability or \(\dfrac{54}{90}\) and It’s more reliable | B0 |
| \(\dfrac{54}{90}\) and There’s a lot of data | B0 |
| Correct ft probability or \(\dfrac{54}{90}\) and He may get better with more throws | B0 |
| \(\dfrac{54}{90}\) and He throws 90 times | B0 |
| Correct ft probability or \(\dfrac{54}{90}\) and More hits | B0 |