Higher June 2025 Paper 3 Q15
15 There are only 7 blue pens and 3 red pens in a box.
Raja takes at random one of the pens.
He notes the colour of the pen and puts the pen back into the box.
Raja does this two more times.
Show that the probability that Raja takes at least two blue pens is \(\dfrac{98}{125}\) (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | for one correct product eg \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{7}{10}\ \left(= \dfrac{343}{1000}\right)\) oe or \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\ \left(= \dfrac{147}{1000}\right)\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{3}{10}\ \left(= \dfrac{27}{1000}\right)\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\ \left(= \dfrac{63}{1000}\right)\) oe or \(\dfrac{7}{10} \times \dfrac{7}{10}\ \left(= \dfrac{49}{100}\right)\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10}\ \left(= \dfrac{9}{100}\right)\) oe |
| M1 | for \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{7}{10}\ \left(= \dfrac{343}{1000}\right)\) oe and \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\ \left(= \dfrac{147}{1000}\right)\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{3}{10}\ \left(= \dfrac{27}{1000}\right)\) oe and \(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\ \left(= \dfrac{63}{1000}\right)\) oe or \(\dfrac{7}{10} \times \dfrac{7}{10}\ \left(= \dfrac{49}{100}\right)\) oe and \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\ \left(= \dfrac{147}{1000}\right)\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10}\ \left(= \dfrac{9}{100}\right)\) oe and \(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\ \left(= \dfrac{63}{1000}\right)\) oe | |
| M1 | for \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{7}{10}\) oe and \(3 \times \dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{3}{10}\) oe and \(3 \times \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\) oe or \(\dfrac{7}{10} \times \dfrac{7}{10}\) oe and \(2 \times \dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10}\) oe or \(\dfrac{3}{10} \times \dfrac{3}{10}\) oe and \(2 \times \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}\) oe | |
| C1 | for a complete method and chain of reasoning leading to \(\dfrac{98}{125}\) eg \(\dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{7}{10} + 3 \times \dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10} = \dfrac{98}{125}\) or \(1 - \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{3}{10} - 3 \times \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10} = \dfrac{98}{125}\) or \(\dfrac{7}{10} \times \dfrac{7}{10} + 2 \times \dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{3}{10} = \dfrac{98}{125}\) or \(1 - \dfrac{3}{10} \times \dfrac{3}{10} - 2 \times \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10} = \dfrac{98}{125}\) |
Additional guidance
Throughout accept probabilities given as decimals or percentages
Condone sampling without replacement for the first method mark only provided it is in the form \(\dfrac{a}{10} \times \dfrac{b}{9} \times \dfrac{c}{8}\) where \(a\), \(b\), \(c\) are integers and \(a \lt 10\) and \(b \lt 9\) and \(c \lt 8\)
Condone any labelling, even if incorrect for the method marks