Higher June 2024 Paper 1 Q13
13 Harman has two bags of beads.
In bag A, there are 3 white beads and 7 black beads.
In bag B, there are 5 white beads and 4 black beads.
Harman takes at random a bead from bag A and a bead from bag B

| Scheme | Marks |
|---|---|
\(\dfrac{3}{10}, \dfrac{7}{10}\) \(\dfrac{5}{9}, \dfrac{4}{9}\) \(\dfrac{5}{9}, \dfrac{4}{9}\) | B2 |
| (2) |
Notes
B2: for all 3 correct pairs of probabilities on the correct branches
If not B2 then award B1 for 1 correct pair of probabilities on a correct branch
Allow equivalent fractions/decimals (to 2 dp truncated or rounded ie 0.55(…) and/or 0.44(…))
| Scheme | Marks |
|---|---|
\(\text{``}{\dfrac{3}{10}}\text{''} \times \text{``}{\dfrac{5}{9}}\text{''}\) oe or \(\text{``}{\dfrac{7}{10}}\text{''} \times \text{``}{\dfrac{4}{9}}\text{''}\) oe or \(\text{``}{\dfrac{3}{10}}\text{''} \times \text{``}{\dfrac{4}{9}}\text{''}\) oe or \(\text{``}{\dfrac{7}{10}}\text{''} \times \text{``}{\dfrac{5}{9}}\text{''}\) oe or | M1ft |
\(\text{``}{\dfrac{3}{10}}\text{''} \times \text{``}{\dfrac{5}{9}}\text{''} + \text{``}{\dfrac{7}{10}}\text{''} \times \text{``}{\dfrac{4}{9}}\text{''}\) oe or \(1 - \left(\text{``}{\dfrac{3}{10}}\text{''} \times \text{``}{\dfrac{4}{9}}\text{''} + \text{``}{\dfrac{7}{10}}\text{''} \times \text{``}{\dfrac{5}{9}}\text{''}\right)\) oe | M1 ft |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{43}{90}\) | A1ft |
| (3) | |
| (5 marks) |
Notes
M1ft: (probabilities < 1)
Allow equivalent fractions/decimals (to 2 dp truncated or rounded ie 0.55(…) and/or 0.44(…))
M1 ft: Allow equivalent fractions/decimals (to 2 dp truncated or rounded ie 0.55(…) and/or 0.44(…))
A1ft: oe
0.47(77..) to 2 dp truncated or rounded or 47.(77)% to 2 sf truncated or rounded