16 She should multiply the number of B1 2 B1 for correct description of process M students (n) she has by the probability of being left-handed (0.14). The number of left-handed students will be 0.14n. 1 17 B1 3 B1 for accuracy and use of Venn oe M P(E only) + + P(M only) = 1 P(E∩M) = P(E∩M) P(E only) = = – Therefore P(M) = 1 P(E only) M1 M1 for use of all probabilities sum to 1 oe – 1 = – M1 M1 for calculation of fraction oe P(Maths) = A1 A1 cao 4 The frequency is approximately the same B1 2 M1 for explanation H 18 for each region of the spinner, suggesting that the spinner is likely to be fair. There are no obvious anomalous results to indicate bias. So there is no strong evidence to suggest it is not fair. 1 19 Joy wins: M1 2 M1 for multiplication to find the number of wins H Vicky’s 0.65 × 52 = 33.8 3 which is approximately 34 wins Vicky won 10 times Joy + Vicky = 44 wins Max = 52-44 = 8 wins Max can expect to win 8 times. A1 A1 cao 2
16 She should multiply the number of students (n) she has by the probability of being left-handed (0.14). The number of left-handed students will be 0.14n. B1 2 B1 for correct description of process M 1 17 P(E only) + P(E∩M) + P(M only) = 1 P(E∩M) = P(E only) = – = Therefore P(M) = 1 – P(E only) 1 – = P(Maths) = B1 M1 M1 A1 3 B1 for accuracy and use of Venn oe M1 for use of all probabilities sum to 1 oe M1 for calculation of fraction oe A1 cao M 4 18 The frequency is approximately the same for each region of the spinner, suggesting that the spinner is likely to be fair. There are no obvious anomalous results to indicate bias. So there is no strong evidence to suggest it is not fair. B1 2 M1 for explanation H 1 19 Joy wins: 0.65 × 52 = 33.8 which is approximately 34 wins Vicky won 10 times Joy + Vicky = 44 wins Max = 52-44 = 8 wins Max can expect to win 8 times. M1 A1 2 3 M1 for multiplication to find the number of Vicky’s wins A1 cao H 2
Foundation_Skills_Probability%20MarkScheme_AQA.docx