Foundation November 2021 Unit 2 Q17
17. Geraint is running a game in a school fair.
A large number of balls are placed in a box.
Each of the balls is one of three colours: bronze, silver or gold.
In the game, a ball is chosen at random from the box.
The table below shows the probability of choosing a bronze ball and the probability of choosing a silver ball.
| Colour | Bronze | Silver | Gold |
|---|---|---|---|
| Probability | 0·68 | 0·22 |
In the game, each person pays £2 to choose a ball at random from the box.
The ball is then returned to the box.
The person wins £3 if a silver ball is chosen.
The person wins £8 if a gold ball is chosen.
There is no prize for choosing a bronze ball.
100 people each play the game once.
How much profit would you expect Geraint to make?
You must show all your working. [5]
| Scheme | Marks | Comments |
|---|---|---|
| ( P(Gold) =) 1 − 0·68 − 0·22 | M1 | |
| = 0·1 | A1 | May be seen in the table. |
| 22 people choose silver AND 10 people choose gold | B1 | FT 100 × ‘their 0·1’. The 10 implies previous M1A1. The 22 and 10 may be seen in further work. |
| (Profit =) 100 × (£)2 – 22 × (£)3 – 10 × (£)8 | M1 | FT ‘their stated number of winners (silver and gold)’. |
| = (£)54 | A1 |
Alternative method1
| Scheme | Marks | Comments |
|---|---|---|
| ( P(Gold) =) 1 − 0·68 − 0·22 | M1 | |
| = 0·1 | A1 | May be seen in the table. |
| 22 people choose silver AND 10 people choose gold | B1 | FT 100 × ‘their 0·1’. The 10 implies previous M1A1. The 22 and 10 may be seen in further work. |
| (Profit =) 68 × (£)2 – 22 × (£)1 – 10 × (£)6 | M1 | FT ‘their stated number of winners (silver and gold)’. |
| = (£)54 | A1 |
Alternative method 2.
| Scheme | Marks | Comments |
|---|---|---|
| ( P(Gold) =) 1 − 0·68 − 0·22 | M1 | |
| = 0·1 | A1 | May be seen in the table. |
| (Profit per game = )(£)2 − 0·22 × (£)3 − 0·1 × (£)8 | M1 | FT ‘their 0·1. |
| = (£)0·54 | A1 | |
| (Total profit = £0·54 × 100 =) (£)54 | B1 | FT ‘their derived £0·54’. |
Alternative method 3.
| Scheme | Marks | Comments |
|---|---|---|
| ( P(Gold) =) 1 − 0·68 − 0·22 | M1 | |
| = 0·1 | A1 | May be seen in the table. |
| (Profit per game = )0·68×(£)2 − 0·22×(£)1 − 0·1×(£)6 | M1 | FT ‘their 0·1. |
| = (£)0·54 | A1 | |
| (Total profit = £0·54 × 100 =) (£)54 | B1 | FT ‘their derived £0·54’. |