Handling Data 2
3 marks · NormalA bag contains 2 yellow counters and 3 green counters. One counter is taken at random, not replaced, and then a second counter is taken at random. The counters are identical except for colour.
List all the possible ordered pairs of counters that could be taken from the bag.
What is the probability that both counters taken are green?
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(±0.01)
(a) YY, YG, GY, GG, YG, GY · must show All 10 ordered pairs: Y1Y2, Y1G1, Y1G2, Y2Y1, Y2G1, Y2G2, G1Y1, G1Y2, G1G2, G2G1
(b) · must show Method for dependent probability: 3/5 × 2/4
Also accept: 3/10; 0.3
Common mistakes: Using 3/5 × 3/5 (as if with replacement).; Incorrectly listing ordered outcomes.
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Marking points
B1 (a) · 1 mark · All possible ordered pairs are listed (10 in total, names or simple notation).
M1 (b) · 1 mark · Chooses dependent probability, 1st green: 3/5, 2nd green: 2/4, then multiply.
A1 (b) · 1 mark · Correct final answer (3/5 × 2/4 = 6/20 = 3/10) or 0.3.
Worked solution
a) Ways to select two counters in order, not replaced:
Let Y = yellow, G = green. Label them Y1, Y2, G1, G2, G3.
Possible ordered pairs: Y1Y2, Y1G1, Y1G2, Y1G3, Y2Y1, Y2G1, Y2G2, Y2G3, G1G2, G1G3, G2G1, G2G3, G3G1, G3G2, ... But since drawing from distinguishable positions, total ways = 5 × 4 = 20. However, as only colour matters, can list as (Y, Y), (Y, G), (G, Y), (G, G), etc; mark as fully as possible (see answer key).
b) P(both green) = probability first counter is green × probability second is green (after a green has been removed):