1MA0 Higher June 2017 Paper 1 Q23
23 There are 7 blue counters, 3 green counters and 1 red counter in a bag.
There are no other counters in the bag.
Hubert takes at random 2 counters from the bag.
(a) Work out the probability that both counters are blue. (3)
(b) Work out the probability that the 2 counters are different colours. (3)
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{42}{110}\) | 3 | M1 for use of 11 and 10 in the denominators M1 for \(\tfrac{7}{11} \times \tfrac{6}{10}\) oe A1 for \(\tfrac{42}{110}\) oe SC for replacement: B1 for \(\tfrac{7}{11} \times \tfrac{7}{11}\ \left(= \tfrac{49}{121}\right)\) |
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{62}{110}\) | 3 | M1 for correct method for GG \(\tfrac{3}{11} \times \tfrac{2}{10}\ \left(= \tfrac{6}{110}\right)\) M1 (dep) \(1 - (\text{BB} + \text{GG}) = 1 - \left(\text{``}\tfrac{42}{110}\text{''} + \text{``}\tfrac{6}{110}\text{''}\right)\) A1 for \(\tfrac{62}{110}\) oe OR M1 for at least two of \(\tfrac{7}{11} \times \tfrac{3}{10}\), \(\tfrac{7}{11} \times \tfrac{1}{10}\), \(\tfrac{3}{11} \times \tfrac{1}{10}\) oe M1 for a complete method eg \(2 \times \left(\tfrac{7}{11} \times \tfrac{3}{10} + \tfrac{7}{11} \times \tfrac{1}{10} + \tfrac{3}{11} \times \tfrac{1}{10}\right)\) oe A1 for \(\tfrac{62}{110}\) oe SC for replacement: B2 for \(2 \times \left(\tfrac{7}{11} \times \tfrac{3}{11} + \tfrac{7}{11} \times \tfrac{1}{11} + \tfrac{3}{11} \times \tfrac{1}{11}\right)\) oe \(\left(= \tfrac{62}{121}\right)\) or \(\left(\tfrac{7}{11} \times \tfrac{4}{11} + \tfrac{3}{11} \times \tfrac{8}{11} + \tfrac{10}{11} \times \tfrac{1}{11}\right)\) oe \(\left(= \tfrac{62}{121}\right)\) or \(1 - \left(\text{``}\tfrac{49}{121}\text{''} + \tfrac{9}{121} + \tfrac{1}{121}\right)\) oe \(\left(= \tfrac{62}{121}\right)\) (B1 for at least two of \(\tfrac{7}{11} \times \tfrac{3}{11}\), \(\tfrac{7}{11} \times \tfrac{1}{11}\), \(\tfrac{3}{11} \times \tfrac{1}{11}\) oe) |