Higher June 2025 Paper 1 Q14
14 There are nine balls labelled 1 to 9 in a box.
Lee will take at random two balls from the box.
Lee says,
“The probability that the sum of the numbers on the two balls will be an even number is greater than the probability that the product of the numbers will be an even number.”
Is Lee correct?
You must show how you get your answer. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| No (supported) | P1 | for P(OO) = \(\dfrac{5}{9} \times \dfrac{4}{8}\ \left(= \dfrac{20}{72}\right)\) or P(OE) = \(\dfrac{5}{9} \times \dfrac{4}{8}\ \left(= \dfrac{20}{72}\right)\) or P(EO) = \(\dfrac{4}{9} \times \dfrac{5}{8}\ \left(= \dfrac{20}{72}\right)\) or P(EE) = \(\dfrac{4}{9} \times \dfrac{3}{8}\ \left(= \dfrac{12}{72}\right)\) |
| P1 | for P(OO) = \(\dfrac{5}{9} \times \dfrac{4}{8}\ \left(= \dfrac{20}{72}\right)\) and P(EE) = \(\dfrac{4}{9} \times \dfrac{3}{8}\ \left(= \dfrac{12}{72}\right)\) OR for P(OE) = \(\dfrac{5}{9} \times \dfrac{4}{8}\ \left(= \dfrac{20}{72}\right)\) and P(EO) = \(\dfrac{4}{9} \times \dfrac{5}{8}\ \left(= \dfrac{20}{72}\right)\) and P(EE) = \(\dfrac{4}{9} \times \dfrac{3}{8}\ \left(= \dfrac{12}{72}\right)\) | |
| P1 | for a process to find probability of sum being even, eg P(OO) + P(EE) = \(\dfrac{5}{9} \times \dfrac{4}{8} + \dfrac{4}{9} \times \dfrac{3}{8}\ \left(= \dfrac{32}{72}\right)\) | |
| P1 | for a process to work with probability of product being even, eg P(EO) + P(OE) + P(EE) = \(\dfrac{4}{9} \times \dfrac{5}{8} + \dfrac{5}{9} \times \dfrac{4}{8} + \dfrac{4}{9} \times \dfrac{3}{8}\ \left(= \dfrac{52}{72}\right)\) or 1 – P(OO) = \(1 - \dfrac{5}{9} \times \dfrac{4}{8}\ \left(= \dfrac{52}{72}\right)\) | |
| C1 | for No supported by correct probabilities, eg \(\dfrac{32}{72}\) and \(\dfrac{52}{72}\) SC B2 for \(\dfrac{41}{81}\) and \(\dfrac{56}{81}\) and No SC B1 for \(\dfrac{41}{81}\) and \(\dfrac{56}{81}\) with no decision or incorrect decision |
Additional guidance
Accept equivalent probabilities throughout
Sample space diagram or listing:
Award P3 for P(sum even) = \(\dfrac{32}{72}\)
or P(product even) = \(\dfrac{52}{72}\), P4 for both