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Students were very confident in answering part (a) with most arriving at the correct value of u.
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Those who understood that P(A or C) represented p + q + s + t were generally able to successfully calculate the value of r , with the majority of students scoring full marks in part (b).
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Those attempting part (c) often gained one of the two method marks available. More commonly this was for realising that q + s = 0.18. To score the second method mark, students needed to use two pieces of information and this proved difficult for many. Had students realised that numerical values could be substituted into their expressions at an early stage, they would have avoided pages of unnecessary algebraic expansion and simultaneous equations. This part proved to be one of the most discriminating parts of the entire paper.
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Most learners appeared comfortable working with tree diagrams and could find the probability of selecting a yellow bead from bag B correctly. A small minority wrote down a correct equation such as 4 3 but did not solve it correctly but the vast majority p x = 12 16 obtained p = 9 and were able to go on and complete the tree diagram correctly. A few 16 responses had different pairs of probabilities on the branches for bag B.
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Most learners appeared comfortable working with tree diagrams and could find the probability of selecting a yellow bead from bag B correctly. A small minority wrote down a correct equation such as 4 3 but did not solve it correctly but the vast majority p x = 12 16 obtained p = 9 and were able to go on and complete the tree diagram correctly. A few 16 responses had different pairs of probabilities on the branches for bag B.
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Part (b) was usually answered well, but the most common error was to overlook the “at least” condition and omit the case for two yellows. The most popular approach was to identify the 4 2 7 ( ) and some used cases from the tree diagram but a few used 1 P 1 Y Y - = - x 3 16 ( ) ( ) 1 P P RR GR - - . Few responses did not present their answers in (b) and (c) as exact fractions as required. (b) or 9 appearing in both the numerator and denominator and leading to an incorrect 16 answer of 1 . 4
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( ) ( ) 1 P P RR GR - - . Few responses did not present their answers in (b) and (c) as exact fractions as required. ( ) P W X ∩ but appreciating that W X meant that the In part (c) many could write down ∩ ( ) P X case GY was required defeated a large number. It was fairly common to see the answer to part
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This question was answered very well. Most knew how to use their calculators to answer part variables and in part (i) calculated P(𝑀𝑀⩾187) whilst others incorrectly applied a continuity (a) though some didn’t appreciate the difference between discrete and continuous random correction. Part (c) was completed successfully by many learners though many used their calculated mean and standard deviation from part (b) to calculate P(170 < F < 175) rather than using the
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Part (a) and part (b) was generally done well by the vast majority of students. A large number of students overcomplicated their workings by calculating probabilities for all 9 possibilities rather than simply calculating P(𝑛𝑜𝑡 7) x P(𝑛𝑜𝑡 7) x P(7). Many students still do not understand the concept of a probability distribution and many attempts were made to calculate a single probability, usually from a binomial distribution, in
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The final question on the paper, as expected, proved to be the most demanding and parts (c) and (d) were only completed by the most able students. part (c). Some students were able to make a good attempt at the probability distribution, but a large number failed to correctly calculate P(𝑆= 4) as they didn't account for the spinner failing to land on a 7 four times. Most students left out part (d) as they were unable to interpret what this probability
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and (d) were only completed by the most able students. Most students left out part (d) as they were unable to interpret what this probability represented. Of those who did try, many complicated probability expressions were seen often leading to an incorrect answer.
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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
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answered part (c) successfully, but some made errors when trying to find P(T < 300) either by using a continuity correction and finding P(T < 299.5) or failing to convert minutes to seconds and trying to find P(T < 5). There were many good attempts at part (d). Most students could start the problem and write
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Part (a) was answered very well with many giving all 3 pairs which was fine provided they were all correct. Part (b) was answered well too with most successfully finding p = 0.13 although a event which was unfortunate as they had usually not made this mistake in part (a).
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all correct. Part (b) was answered well too with most successfully finding p = 0.13 although a few students omitted the 0.08 from P(C). Part (c) proved more discriminating. Whilst many
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few students omitted the 0.08 from P(C). Part (c) proved more discriminating. Whilst many ( ) P A B students knew ( ) = P | A B they were not able to find the relevant values from the ( ) P B Venn diagram. Some thought ( ) ( ) ( ) = P P P A B A B and quite a few were unable to establish the relationship q + r = 0.22 and some of those who did get this far believed that r couldn’t be 0 and gave an answer of 0.21 . A common error was to write 0.55 ( ) P A + 0.05 q ( ) = = P | A B . In part (d), having been given an equation with ( ) + + + P 0.25 0.08 B q r
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( ) P A + 0.05 q ( ) = = P | A B . In part (d), having been given an equation with ( ) + + + P 0.25 0.08 B q r which to work, students were much more successful with many gaining all 3 marks. There were errors with the conditional probability again and many slips in adding probabilities, but it was 0.4 r = 0.5 common to see leading to the correct values for r and q. When part (e) was + 0.73
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0.4 r = 0.5 common to see leading to the correct values for r and q. When part (e) was + 0.73 attempted, it was usually correct however a few students multiplied 0.25 and 0.08 instead of adding them. Part (f) proved to be one of the most discriminating parts of the paper. There were
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adding them. Part (f) proved to be one of the most discriminating parts of the paper. There were many blank responses, and some simply gave an answer of p = 0.13. Some though could find the correct region but gave the answer as a probability e.g. ( ) P B A C rather than an event which was unfortunate as they had usually not made this mistake in part (a).
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This was the second most demanding question on the paper. In this question there was some confusion between the number of students who study Art and the number of students who study Art but not Music. This often led to incorrect Venn diagrams and subsequent errors in both parts of the question. In part (a), students often found that at most 20 students studied Music but not Art, but then went no further. Very few were able to find both required end-points.
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This was the second most demanding question on the paper. In this question there was some confusion between the number of students who study Art and the number of students who study Art but not Music. This often led to incorrect Venn diagrams and subsequent errors in both parts of the question. The rule for independence is given to students on the formulae booklet, but many quoted it in terms of A and B rather than the events A and M as required in part (b) of this question. Students making little progress often tended to use numbers rather than probabilities throughout. Only a small minority arrived at the correct result of x = 21.6, but some then concluded that A and M were independent as 21.6 was in the range, ignoring the fact that x had to be an integer in this context.
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The final question on the paper, as expected, proved to be the most demanding as many students were unable to make any progress at all here. Part (a) was the most discriminating with many students leaving this part blank. Students often thought that X followed a binomial distribution. Some tried to work out individual probabilities for each letter rather than focusing on the number of A’s that could be selected. A large number of tree diagrams were seen but some could not get from the diagram to a probability distribution as they were unable to give a probability for each of the different values of X. Students who correctly worked out the probability of obtaining 1 or 2 A's sometimes did not recognise that the answer needed multiplying by 3.
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This proved to be a positive and accessible start to the paper. Part (a) was nearly always correct and most knew how to write down a suitable expression for independence for part (b) which usually resulted in a correct equation for p. Those
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This proved to be a positive and accessible start to the paper. for independence for part (b) which usually resulted in a correct equation for p. Those that didn’t achieve this often thought that P(C) = 0.3 or 0.3p or even just p. Solving for p was mostly completed correctly and even those who had an incorrect value for p could often score the follow through for q provided their p + q = 0.27
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This proved to be a positive and accessible start to the paper. In part (c), some used the simple conditional probability formula = [ P( ] in the formula booklet to get started whereas others just wrote A ) P( )P( | ) B A A B down a correct ratio of probabilities from their Venn diagram. A fairly common error was to have P(B) = 0.6 as the denominator and a few misinterpreted the formula and used ( ) P A B . Those using the more complicated formula for P(A|B) in the formula booklet ( ) P B rarely managed to complete part (c) and students should be encouraged to know how to use the formulae in the formula booklet efficiently.
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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
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This question proved to be the most challenging on the paper for students.
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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.
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This question also proved to be quite challenging.
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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.
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Part (a) was answered well by most candidates but often there was little explanation offered to explain where their probabilities came from, nevertheless the correct answers to (i) and (ii) were + but they had at often seen. Some thought that, for example, in (i) the answer should be 3 3 1 1 4 3 3 4 least realised that the red spinner must have landed on 3 and the green spinner on 4 and scored the first mark. Part (b) was designed as a problem solving question and it was good to see a number of
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first mark. Part (b) was designed as a problem solving question and it was good to see a number of students trying it though understandable that many felt there were easier pickings available in paper 22 and moved onto that. Most attempts involved setting up two linear equations and solving them but there was some confusion over values of the random variables and probabilities associated with these values. However over 10% of candidates scored full marks on this question and shoed a good understanding of this topic.
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Part (a) was well answered with most students able to gain the mark for a correct answer. A simplified fraction or the correct decimal value was often seen. Whilst the majority of students continued to score well in part (b) the most common
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Whilst the majority of students continued to score well in part (b) the most common incorrect answer came from students who thought the question was conditional, giving 170 550 , i.e. the probability of not being a professional given they live in area B. Some students attempted a product of probabilities assuming independence, which was not given and not correct.
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In part (c) there was mixed success in the interpretation of the information given in the question with only about half of students getting the Venn diagram completely correct. Some correct diagrams followed minimal working whilst others filled at least a page. A ′∩′∩ ′ blank. The method mark was common omission was to leave the region H R F most commonly given when implied by one correct value in the Venn diagram, rather than for the student’s working, which was commonly unclear. Some working indicated incorrect use of the given percentages. Where only one value was correct in the Venn diagram it was usually the 481 or 259, for ‘professionals from area A’. When none or only one of the four correct values was seen many students benefited from being able to achieve 1 mark for the sums of certain pairs of values in their diagram. There were a number of students who simply abandoned the Venn diagram which often resulted in no attempts being made at the remaining parts of the question.
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Part (d) had a mixed response with many correct answers. This part could be answered independently of whether or not (c) had been attempted and it was good to see most students doing so successfully through mark could only be scored by students who had some value in the region 𝐻𝐻′ ∩ 𝑅𝑅′ ∩𝐹𝐹′ so that this value could be added to the 133 in their calculation. A common Whilst many realised the 133 given in the Venn diagram was required, the follow- non-scoring answer of 133 1825 was often seen. Most students realised that part (d) was a conditional probability question with many taking the simpler approach of using numbers rather than probabilities in their ratio. Those with a correct Venn diagram usually scored both marks here. As is often the case, a noticeable number of students had the incorrect event, namely F, as the denominator in their probability expression rather than H.
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Part (a) was well answered with most students able to gain the mark for a correct answer. A simplified fraction or the correct decimal value was often seen. Whilst the majority of students continued to score well in part (b) the most common incorrect answer came from students who thought the question was conditional, giving 170 550 , i.e. the probability of not being a professional given they live in area B. Some students attempted a product of probabilities assuming independence, which was not given and not correct. In part (c) there was mixed success in the interpretation of the information given in the question with only about half of students getting the Venn diagram completely correct. Some correct diagrams followed minimal working whilst others filled at least a page. A ′∩′∩ ′ blank. The method mark was common omission was to leave the region H R F most commonly given when implied by one correct value in the Venn diagram, rather than for the student’s working, which was commonly unclear. Some working indicated incorrect use of the given percentages. Where only one value was correct in the Venn diagram it was usually the 481 or 259, for ‘professionals from area A’. When none or only one of the four correct values was seen many students benefited from being able to achieve 1 mark for the sums of certain pairs of values in their diagram. There were a number of students who simply abandoned the Venn diagram which often resulted in no attempts being made at the remaining parts of the question. Part (d) had a mixed response with many correct answers. This part could be answered independently of whether or not (c) had been attempted and it was good to see most students doing so successfully through mark could only be scored by students who had some value in the region 𝐻𝐻′ ∩ 𝑅𝑅′ ∩𝐹𝐹′ so that this value could be added to the 133 in their calculation. A common Whilst many realised the 133 given in the Venn diagram was required, the follow- non-scoring answer of 133 1825 was often seen. Most students realised that part (d) was a conditional probability question with many taking the simpler approach of using numbers rather than probabilities in their ratio. Those with a correct Venn diagram usually scored both marks here. As is often the case, a noticeable number of students had the incorrect event, namely F, as the denominator in their probability expression rather than H.
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Part (a) was well answered with most students able to gain the mark for a correct answer. A simplified fraction or the correct decimal value was often seen. Whilst the majority of students continued to score well in part (b) the most common incorrect answer came from students who thought the question was conditional, giving 170 550 , i.e. the probability of not being a professional given they live in area B. Some students attempted a product of probabilities assuming independence, which was not given and not correct. In part (c) there was mixed success in the interpretation of the information given in the question with only about half of students getting the Venn diagram completely correct. Some correct diagrams followed minimal working whilst others filled at least a page. A ′∩′∩ ′ blank. The method mark was common omission was to leave the region H R F most commonly given when implied by one correct value in the Venn diagram, rather than for the student’s working, which was commonly unclear. Some working indicated incorrect use of the given percentages. Where only one value was correct in the Venn diagram it was usually the 481 or 259, for ‘professionals from area A’. When none or only one of the four correct values was seen many students benefited from being able to achieve 1 mark for the sums of certain pairs of values in their diagram. There were a number of students who simply abandoned the Venn diagram which often resulted in no attempts being made at the remaining parts of the question. Part (d) had a mixed response with many correct answers. This part could be answered independently of whether or not (c) had been attempted and it was good to see most students doing so successfully through mark could only be scored by students who had some value in the region 𝐻𝐻′ ∩ 𝑅𝑅′ ∩𝐹𝐹′ so that this value could be added to the 133 in their calculation. A common Whilst many realised the 133 given in the Venn diagram was required, the follow- non-scoring answer of 133 1825 was often seen. Most students realised that part (d) was a conditional probability question with many taking the simpler approach of using numbers rather than probabilities in their ratio. Those with a correct Venn diagram usually scored both marks here. As is often the case, a noticeable number of students had the incorrect event, namely F, as the denominator in their probability expression rather than H.
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The vast majority were able to find the correct value for p but under half managed to identify a pair of mutually exclusive events. Some lost the mark for confusing events with probabilities of events and others thought the answer was A and B or B and C.
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The vast majority were able to find the correct value for p but under half managed to identify a pair of mutually exclusive events. Some lost the mark for confusing events with probabilities of events and others thought the answer was A and B or B and C.
Sadly two thirds of the candidates were not able to make any progress with this question. Those who did pause for a moment and consider the situation often realised that there were two possible cases to consider. The majority of these failed to give the correct probabilities though as they did not realise that after the initial transfer from bag A there were 10, rather than 9, balls in bag B. A handful of the students did negotiate the problem successfully though some then added together their probabilities (instead of giving the two pairings of n and p) and lost the final mark.
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of 0.022 here and lost this mark. Part (c) was usually answered correctly and many went on to use the correct binomial distribution in part (d) though some found interpreting “fewer than 3” a problem and a
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correct binomial distribution in part (d) though some found interpreting “fewer than 3” a problem and a ). Overall though this proved a good start to the paper common incorrect answer was 0.924 (from P( X 3 and over 40% scored 6 or 7 marks here.
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This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% scoring full marks. Most students navigated parts (a) and (b) successfully but a few missed the information that “No students read all three magazines” and wasted a lot of time trying to find two equations in p and q. The conditional probability in part (c) proved to be a challenge for some; there were those who effectively
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This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% scoring full marks. Most students navigated parts (a) and (b) successfully but a few missed the information that “No students read all three magazines” and wasted a lot of time trying to find two equations in p and q. The conditional probability in part (c) proved to be a challenge for some; there were those who effectively
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This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% The conditional probability in part (c) proved to be a challenge for some; there were those who effectively used P(E | S) and others were miscopying values from their Venn diagram. Those who failed to make any progress in (i) were often able to pick up a mark for a correct follow through when finding the value of t.
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This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% Part (d) was the least successful part of the question often due to poor use of the notation. There were a curiously large number of students who confused events with the probability of events; for example we saw E′ E′ a number of cases of students writing S = 0.48 instead of P( S ) = 0.48. Another source of errors ∩ ∩ and was confusing the symbols ∪ in their expressions. There was a demand to show clear working here ∩ and this meant that we needed to see probabilities correctly labelled and an appropriate test clearly shown and a number of attempts lacked these features but there were plenty of good solutions to this part too.
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this question. Identifying the conditional probability in part (c) defeated many students and some of those who seemed to be attempting to use a conditional probability were unable to use the correct notation to write their expression. Those who did start with a correct conditional probability and moved onto a ratio of probabilities often had a correct denominator but a common error was to have P(F > 160) as the numerator rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something
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students found the correct set of values with varying degrees of explanation on the way. Part (b) was often left blank though the first mark was available without values for a, b and c provided brackets were used correctly for the expression. Nearly 10% managed to score full marks for the question.
There were some very good solutions given to this question with clear and organised working but there were many others with a correct statement lurking amid a pile of detritus. Candidates need to plan a course of action before writing numbers down. The most common error was to double the wrong set of probabilities, but there were still marks to be gained in further working. There were too many candidates who did not see the need to use the sum of probabilities as part of a solution. The candidates who reached the latter stages generally found the correct ways in which to obtain a total of 6. One of the more frequent errors in finding the total of 6 was think that there are two ways of arranging 3,3.
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This proved to be a popular start to the paper with over 35% of students scoring full marks. Part (a) was answered well with most understanding the term “mutually exclusive” and giving their answer as events not probabilities. Part (b) was very
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This proved to be a popular start to the paper with over 35% of students scoring full exclusive” and giving their answer as events not probabilities. Part (b) was very straightforward and nearly all could interpret the Venn diagram correctly to answer this. Many knew a suitable rule for independence in part (c) and could use this to find P(A) =
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This proved to be a popular start to the paper with over 35% of students scoring full Many knew a suitable rule for independence in part (c) and could use this to find P(A) = 0.6 or write down a suitable equation for q which was then solved correctly. Part (d)
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This proved to be a popular start to the paper with over 35% of students scoring full 0.6 or write down a suitable equation for q which was then solved correctly. Part (d) was more challenging as many could not use the given conditional probability correctly to obtain an equation for r but most knew how to use this answer to find s and were able to gain at least one mark in this part.
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identify some, though rarely all, the cases in (b). Finding the probabilities was more of a problem with some using P(D = 50) + P(D = 30) rather than the product and those having correct probabilities for all 3 cases were quite rare. It was good to see many students who had faltered on (b) still making a start on part (c). Many identified the 4
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It was disappointing that over 20% of candidates scored no marks here as there were some straightforward marks available to those who had studied this part of the In (c)(ii) although a good number of students recognised that a conditional probability was required, few could find a correct expression for the numerator and a common error (c)(i) was to see used rather than a correct numerator of P(0 < T < 2). In part (iii) P( 0) T > many realised that the model would assign some probability to negative values of T but this will always be true when a normal distribution is used as a model. The point here was that a significant proportion of the values for T were negative and so this means that the normal model is not very good. Most responses seemed to believe that a model must match reality exactly but a few did comment that P(T > 0) should be closer to 1 or that the difference between (i) and (ii) was too big and therefore the model was not a good one to use. Fully correct answers to part (d) were rare and this part gave the A* students an
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It was disappointing that over 20% of candidates scored no marks here as there were some straightforward marks available to those who had studied this part of the Fully correct answers to part (d) were rare and this part gave the A* students an opportunity to shine. Successful students realised that they needed to look at P(T > 2) and often found this by taking 1 – (c)(i) but they rarely realised that the median splits this probability into 2 so they were seeking the value of t such that P(T > t) = 0.40…Some who reached this stage tried, incorrectly to use their calculator, failing to realise that their calculator gave t for probabilities of the form P(T < t), a few used the tables provided to find z = 0.2533 and some did prepare correctly to use their calculator and found P(T < t ) = 0.5978… (the 0.402… + (c)(i)) from whence the correct value of t = 5.9 could be found. It was encouraging to see some good use of diagrams in this question being used to help students identify which probabilities they needed.
Only around 10% of candidates achieved full marks on this question, but there were many strong attempts made. Most scored the first mark for identifying that x = 0. The most common error across the question was to ignore the intersections and calculate P(A) x P(C) as 0.1 x 0.39. The majority of candidates scored the final method mark for a correct equation in y and z, demonstrating knowledge of probabilities summing to 1. Surprisingly, of candidates who did set up all of the equations required, many were unable to manipulate the algebra to solve them. Those who confused independent and mutually exclusive events tended to make no progress.
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On the whole, this accessible question was answered very well by majority of candidates with nearly 40% going on to achieve full marks. Most were able to draw the correct tree diagram in part (a) and label both outcomes and probabilities clearly. However, a number of candidates made mistakes due to deciding Bag A contained only 9 marbles, B contained 4 marbles and C contained 2 marbles. Some candidates attempted a 3 branch tree diagram and made little progress. Others thought that the successive draws from Bag C could take place. All branches must be labelled as blank branches are not assumed to represent a probability of 0.
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On the whole, this accessible question was answered very well by majority of candidates with nearly 40% going on to achieve full marks. Most were able to draw the correct tree diagram in Part (b) was almost universally answered correctly, even if the tree diagram was incorrect, with the majority of candidates gaining full marks or at least follow through marks, although occasionally the three probabilities were added rather than multiplied together. Many correct responses were seen in part (c) with the majority of candidates gaining full marks
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On the whole, this accessible question was answered very well by majority of candidates with nearly 40% going on to achieve full marks. Most were able to draw the correct tree diagram in Many correct responses were seen in part (c) with the majority of candidates gaining full marks or at least follow through marks for the method. The most common error was the inclusion of 0.1 for P(red) in addition to the two correct products, perhaps indicating that the candidates failed to interpret ‘at least’ correctly.
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On the whole, this accessible question was answered very well by majority of candidates with nearly 40% going on to achieve full marks. Most were able to draw the correct tree diagram in Part (d) was somewhat less successfully answered. Many candidates were unable to recognise 9 1 that this involved conditional probability, with the most common wrong answer of 5 often 10 x seen. A small number had the conditional probability the wrong way round, attempting to find P(R|B) instead of P(B|R).
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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 context what these mean. Most earned the mark in part (a) except the few candidates who did not give an answer to the required degree of accuracy.
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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 Most who attempted part (d) scored the mark. was to compare the result in (d) to 0.76. The failure to focus on the fact that the model was Binomial, and whether therefore the conditions of independence had been met, demonstrated the general unfamiliarity with modelling.
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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
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Despite the large number of students scoring zero here, about 1/3 managed to answer part (a) successfully. Part (b) required an explanation which means both words and a simple calculation were
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successfully. Part (b) required an explanation which means both words and a simple calculation were required. Some found P(B) x P(F) = 0.018 and others gave the probability of the component coming from B and being faulty as 0.009 but both of these calculations and a concluding statement were needed to secure the mark.
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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
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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.
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Whilst parts (a) and (d) were often answered well, parts (b) and (c) proved much more challenging and only the strongest 10% made significant progress here. Part (a) was answered well and most identified that a conditional probability was required in part (b) were often able to make progress, often obtaining the probability of 0.4462 and using that correctly to complete part (b) and make progress in (c).
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Whilst parts (a) and (d) were often answered well, parts (b) and (c) proved much more challenging and only the strongest 10% made significant progress here. Part (a) was answered well and most progress in (c). In part (c), some simply squared their probability of 0.4462 without realising the need to consider the two new batteries as well. Part (d) was often answered quite well with many stating the hypotheses
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