S3.3 Modelling with probability

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was answered very well by most students and in part (b) a good proportion also recognised the need for the events to be independent. Some students failed to score the mark in part (b) because they did not use correct mathematical language: we saw comments such as “they are not affected by each other” or “the events are not mutually exclusive” and a few simply said “it is continuous data” or “they must take part in both events”. Many students answered part (c) successfully, but some made errors when trying to find P(T < 300) either by

Common mistakes

• H and T (or the two events) must be independent
• "not mutually exclusive", "not affected by each other" or "the results are independent" not accepted

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be the most challenging on the paper for students. In part (a) few could see how to use the two bullet points to reach the required result. A k good number realised that P(S |{X = x}) = x but couldn’t always use the conditional k probability formula to reach ( ) = = . Those that did get this far were P 50 S X b 50 often able to give a convincing proof of the given result. A common misconception was k k = = believing that and then using the sum of the probabilities equalling 1 , etc a b 20 50 but this scored no marks. It was good to see a number of students using the result from part (a) to form similar relationships between the probabilities, such as d = 2b and going on to solve their equations to find the probability distribution. The final part was particularly challenging with many misinterpreting the 30% probability

Common mistakes

• P(S ∩ X = x) = P(S given X = x) × P(X = x), so b × k/50 = c × k/80
• assuming k/50 = b (and summing to 1) scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be the most challenging on the paper for students.

Common mistakes

• form further relations (d = 2b, a = 2b/5) from the same idea, then sum of probabilities = 1: b = 1/5, a = 2/25, c = 8/25, d = 2/5

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be the most challenging on the paper for students. The final part was particularly challenging with many misinterpreting the 30% probability given in the stem to part (c). Many confused this 0.3 [= P(S | X = 100) ] with the 0.4 [= P(X = 100)] from the probability distribution and simply compared these two values. Those that did realise that Nav’s 30% meant that k = 30 often went on to show that when x = 20 this would give a probability [P(S | X = 20)] > 1 and so the model was not suitable.

Common mistakes

• 0.3 is P(S given X = 100), so k = 30
• then P(S given X = 20) = 1.5 > 1, impossible
• common error comparing 0.3 with P(X = 100) = 0.4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question also proved to be quite challenging. find the required probability. In part (f) only a small number of candidates realised that the fact that some patients might stay longer than 40 hours was a limitation of the model. Others did not give any context in their answers or discussed the shape of the curve without considering the stated upper limit.

Common mistakes

• model only defined for 0 ≤ x ≤ 4, so patients staying longer than 40 hours are not covered
• answer must be in context, not about the shape of the curve

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question demonstrated that most candidates are confident in carrying out statistical calculations, particularly probabilities, but only the most able are confident in interpreting in Part (e) was the least successfully answered part of the question with lots of blanks seen. Where candidates scored the 2 marks it was usually through stating that if there had been cloud cover the day before there was a higher chance of cloud cover the next day. Those going completely down the wrong route failed to realise that the data was from the same source and were mostly commenting on the size of the sample rather than what the statistics already demonstrated. In general there was far too much tendency to describe vague contextual knowledge rather than use the figures found previously. Of those who did try and compare figures an error often seen was to compare the result in (d) to 0.76. The failure to focus on the fact that the model was

Common mistakes

• 0.821 is higher than 0.717 – high cloud cover is more likely after a day of high cover, so independence fails and the binomial model may not be suitable
• comparing with 0.76 or commenting on sample size is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) was answered fairly well, but in part (c) many carried on using the B(10, 0.1) model finding P(H = 5) rather than P(F = 5). Part (d) proved more challenging; those who carefully started to list P(F = 5) had to equal their answer to part (c) and made little progress in (d) but could often collect the mark in (e). Part (f) was not answered well; the question clearly steered students to comment on the

Common mistakes

• first hit on 5th throw: 0.9⁴ × 0.1 = 0.0656
• common error: continuing with B(10, 0.1) to find P(H = 5)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

mark in (e). Part (f) was not answered well; the question clearly steered students to comment on the probability of a dart hitting the target using these two models. There were some good descriptions of what was happening with Thomas’ model but often these were not compared to Peta’s model.

Common mistakes

• Peta: probability of hitting constant
• Thomas: probability increases with each attempt
• must compare both models, not describe Thomas' alone