Higher June 2019 Paper 1 Q13
13 In group C, there are 6 girls and 8 boys.
In group D, there are 3 girls and 7 boys.
A team is made by picking at random one child from group C and one child from group D.
(a) Complete the probability tree diagram. (2)
(b) Work out the probability that there are two boys in the team. (2)
After the first team has been picked, a second team is picked.
One child is picked at random from the children left in group C and one child is picked at random from the children left in group D.
(c) Work out the probability that there are two boys in each of the two teams. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{6}{14}, \dfrac{8}{14}\) | B1 |
| \(\dfrac{3}{10}, \dfrac{7}{10}, \dfrac{3}{10}, \dfrac{7}{10}\) | B1oe |
| (2) |
Notes
B1: for \(\dfrac{6}{14}\left(\dfrac{3}{7}\right), \dfrac{8}{14}\left(\dfrac{4}{7}\right)\) in correct positions. Allow decimals of 2dp or better (0.43, 0.57)
B1oe: for \(\dfrac{3}{10}, \dfrac{7}{10}, \dfrac{3}{10}, \dfrac{7}{10}\) in correct positions.
| Scheme | Marks |
|---|---|
| \(\dfrac{8}{14} \times \dfrac{7}{10}\) | M1 |
| \(\dfrac{2}{5}\) oe | A1 |
| (2) |
Notes
M1: ft from (a)
| Scheme | Marks |
|---|---|
\(\dfrac{7}{13} \times \dfrac{6}{9}\left(= \dfrac{42}{117} = \dfrac{14}{39} = 0.35(897\ldots)\right)\) or \(\dfrac{8}{14} \times \dfrac{7}{13}\left(= \dfrac{56}{182}\text{ oe}\right)\) or \(\dfrac{7}{10} \times \dfrac{6}{9}\left(= \dfrac{42}{90}\right)\) | M1 |
| \(\text{``}{\dfrac{42}{117}}\text{''} \times \text{``}{\dfrac{2}{5}}\text{''}\) or \(\left(\dfrac{8}{14} \times \dfrac{7}{13}\right) \times \left(\dfrac{7}{10} \times \dfrac{6}{9}\right)\) | M1 |
| \(\dfrac{28}{195}\) oe | A1 |
| (3) | |
| (7 marks) |
Notes
M1: ft from (a)
(\(\dfrac{7}{13} = 0.54\) to 2dp
\(\dfrac{6}{9} = 0.67\) to 2dp)
M1: ft from (b)
A1: for \(\dfrac{28}{195}\) oe, e.g. 0.14(3589…) from accurate working