and Progress descriptor 4a 2k + + 0 + = 1 M1 2.1 4th k k Calculate A1* 1.1b 4k = 1, so = 0.25 (answer given). k probabilities from discrete distributions. (2) 4b P(X1 + = 5) = P(X1 = 3 and = 2) + P(X1 = 2 and = 3) B1* 2.4 4th X2 X2 X2 = 0 + 0 = 0 (answer given). Calculate probabilities from discrete distributions. (1) 4c M1 2.5 4th A1 1.1b Calculate probabilities from A1 1.1b discrete distributions. (3) 4d P(1.3 + 3.2) = P(X1 + = 2) + P(X1 + = 3) M1 3.4 4th X1 X2 X2 X2 ⩽ ⩽ Calculate probabilities from discrete = 0.0625 + 0.25 = 0.3125 or A1ft 1.1b distributions. (2) (8 marks) 4b Must show that 5 can only be obtained from 2 and 3 or 3 and 2, and so must use P(X = 2) = 0 but condone explanation in words. 4c M1 for correct set of values for + Condone omission of 5 column. X1 X2. A1 for correct probabilities for 0, 2 and 6. A1 for others. Equivalent fractions are Pearson Q Scheme Marks AOs Progression Step
and Progress descriptor 4a 2k + k + 0 + k = 1 M1 2.1 4th Calculate probabilities from discrete distributions. 4k = 1, so k = 0.25 (answer given). A1* 1.1b (2) 4b P(X1 + X2 = 5) = P(X1 = 3 and X2 = 2) + P(X1 = 2 and X2 = 3) = 0 + 0 = 0 (answer given). B1* 2.4 4th Calculate probabilities from discrete distributions. (1) 4c M1 A1 A1 2.5 1.1b 1.1b 4th Calculate probabilities from discrete distributions. (3) 4d P(1.3 ⩽ X1 + X2 ⩽ 3.2) = P(X1 + X2 = 2) + P(X1 + X2 = 3) M1 3.4 4th Calculate probabilities from discrete distributions. = 0.0625 + 0.25 = 0.3125 or A1ft 1.1b (2) (8 marks) 4b Must show that 5 can only be obtained from 2 and 3 or 3 and 2, and so must use P(X = 2) = 0 but condone explanation in words. 4c M1 for correct set of values for X1 + X2. Condone omission of 5 column. A1 for correct probabilities for 0, 2 and 6. A1 for others. Equivalent fractions are Q Scheme Marks AOs Pearson Progression Step
14_as_statistics_and_mechanics_practice_paper_g_mark_scheme.docx