Higher November 2023 Unit 1 Q21
21.
(a) A box contains seven black counters, three white counters and one red counter.
Aled takes two counters at random from the box. These counters are not replaced.
Calculate the probability that the two counters that Aled chose are both the same colour. [3]
Aled takes two counters at random from the box. These counters are not replaced.
Calculate the probability that the two counters that Aled chose are both the same colour. [3]
(b) A second box contains \(n\) yellow cards and \((n + 1)\) red cards.
Delyth takes two cards at random from the second box. These cards are not replaced.
What is the probability that the two cards that Delyth chose are both yellow?
Give your answer as an algebraic fraction in its simplest form. [3]
Delyth takes two cards at random from the second box. These cards are not replaced.
What is the probability that the two cards that Delyth chose are both yellow?
Give your answer as an algebraic fraction in its simplest form. [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{7}{11} \times \dfrac{6}{10} + \dfrac{3}{11} \times \dfrac{2}{10} + \dfrac{1}{11} \times \dfrac{0}{10}\) | M2 | Full method for finding P(B,B) + P(W,W) + P(R,R). May be seen in parts. Allow the omission of P(R,R). M1 for sight of \(\dfrac{7}{11} \times \dfrac{6}{10}\) or \(\dfrac{3}{11} \times \dfrac{2}{10}\). M1 if additional incorrect products are included. |
| = \(\dfrac{48}{110}\ \left(= \dfrac{24}{55}\right)\) | A1 | ISW If no marks, SC1 for one of the following:
|
Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| \(1 - \left(\dfrac{7}{11} \times \dfrac{4}{10} + \dfrac{3}{11} \times \dfrac{8}{10} + \dfrac{1}{11} \times \dfrac{10}{10}\right)\) | M2 | Full method for finding 1 – P(BBʹ or WWʹ or RRʹ) M1 if one of the subtracted products is incorrect or missing. |
| = \(\dfrac{48}{110}\ \left(= \dfrac{24}{55}\right)\) | A1 | ISW If no marks, SC1 for one of the following:
|
Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| \(1 - 2 \times \left(\dfrac{7}{11} \times \dfrac{3}{10} + \dfrac{3}{11} \times \dfrac{1}{10} + \dfrac{1}{11} \times \dfrac{7}{10}\right)\) or equivalent | M2 | Full method for finding 1 – P(BW or WB or WR or RW or RB or BR) M1 if one or two of the subtracted products are incorrect or missing. |
| = \(\dfrac{48}{110}\ \left(= \dfrac{24}{55}\right)\) | A1 | ISW If no marks, SC1 for one of the following:
|
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{n}{2n+1} \times \dfrac{n - 1}{2n}\) or equivalent | M2 | M1 for sight of \(\dfrac{n}{2n+1}\) or \(\dfrac{n - 1}{2n}\) or equivalent For M1, FT ‘their \(2n+1\)’– 1 in denominator of 2nd fraction. |
| \(= \dfrac{n-1}{2(2n+1)}\) or \(\dfrac{n-1}{4n+2}\) | A1 | CAO |