(a) Rita has a box containing 12 buttons, 5 are white (W) and the rest are blue (B). Rita takes 2 buttons from the box at random, without replacement.
Complete the tree diagram. [3]
(b) Calculate the probability that at least one of the buttons Rita takes from the box is white. [2]
(c) Rita puts the 2 buttons back into the box. She then takes 3 buttons from the box at random, without replacement.
What is the probability that the second and the third buttons she takes are both blue? [3]
Mark scheme (a)
Scheme
Marks
Comments
Correct tree diagram e.g.
B3
B1 for left hand branches correct FT ‘their \(\dfrac{5}{12}\) or \(\dfrac{7}{12}\)’ B1 FT for top right branches correct B1 FT for bottom right branches correct
Allow equivalent fractions in all cases.
If rounded decimals used here e.g. 0.42, 0.58, 0.36, 0.64, 0.45 and 0.55 award B1 here and FT in (b) and (c) for M marks only.
FT ‘their probabilities’ providing at least B1 awarded in (a)
\(\dfrac{90}{132}\) or \(\dfrac{15}{22}\) oe
A1
FT ‘their tree diagram’
If no marks award SC1 for the correct sum of two of the required products e.g for: \(\frac{5}{12} \times \frac{7}{11} + \frac{7}{12} \times \frac{5}{11} = \)\(\frac{70}{132}\) oe
Corrected by Westie's Workshop
The board's mark scheme printed “\(\frac{70}{108}\)”. Changed to “\(\frac{70}{132}\)” because \(\frac{5}{12} \times \frac{7}{11} + \frac{7}{12} \times \frac{5}{11} = \frac{35}{132} + \frac{35}{132} = \frac{70}{132}\).
Original: see the board's mark scheme PDF (link above).
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