Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.
- (i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
- (ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
Step-by-step solution
Idea: Because the pen is put back, both picks are from the same 9 pens. The colours are not equally likely (black has the most pens), so put the probabilities on the branches and multiply along each path.
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
- Your pick: 3 branches. P(red) = 39, P(black) = 49, P(green) = 29.½ mark
- Friend’s pick: the pen was put back, so from each of your branches draw the same 3 branches with the same probabilities 39, 49, 29.½ mark
- Possible colour outcomes (your colour first): {RR, RB, RG, BR, BB, BG, GR, GB, GG}: 3 × 3 = 9 outcomes, shown at the ends of the 9 paths of the tree.1 mark
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
- The 9 colour outcomes are not equally likely, so do not just write 39. Multiply along each same-colour path:
P(RR) = 39 × 39 = 981, P(BB) = 49 × 49 = 1681, P(GG) = 29 × 29 = 481.1 mark - P(same colour) = 981 + 1681 + 481 = 2981 ≈ 0.36.
(Counting check: there are 9 × 9 = 81 equally likely (your pen, friend’s pen) pairs; same colour: 3 × 3 + 4 × 4 + 2 × 2 = 29 of them.)1 mark
Check: All 9 path probabilities add to 1: (9 + 12 + 6 + 12 + 16 + 8 + 6 + 8 + 4)/81 = 8181 ✓. So P(different colours) = 5281 and 2981 + 5281 = 1 ✓.
Answer to write in the exam
(i)
Each pick: P(R) = 39, P(B) = 49, P(G) = 29 (pen replaced)
Tree: 3 branches, each followed by 3 branches (as drawn)
∴ Outcomes = {RR, RB, RG, BR, BB, BG, GR, GB, GG}
(ii)
P(RR) = 39 × 39 = 981
P(BB) = 49 × 49 = 1681
P(GG) = 29 × 29 = 481
∴ P(same colour) = 981 + 1681 + 481 = 2981 ≈ 0.36
Common mistakes that cost marks
- Guessing 39 = 13 because 3 of the 9 colour outcomes are ‘same colour’. The colour outcomes are not equally likely; BB is four times as likely as GG.
- Using 38 or other ‘8’ fractions for the friend’s pick. The pen is put back, so the friend still picks from 9 pens.
- Adding along a path (39 + 39) instead of multiplying (39 × 39).
How this can come in the exam
A bag has 2 red and 3 white balls. A ball is drawn, put back, and a second ball is drawn. The probability that both are white is
- 35
- 625
- 925
- 310
Show answer
(C) 925
35 × 35 = 925 (with replacement, 5 × 5 = 25 equally likely pairs, 3 × 3 = 9 favourable).
A box has 1 yellow and 2 blue chalks. Sita picks one, puts it back, then Gita picks one. Draw a tree diagram and find the probability that they pick different colours.
Show answer
Each pick: Y 13, B 23. Paths: YY 19, YB 29, BY 29, BB 49. Different colours: YB + BY = 29 + 29 = 49.Try one yourself
A box has 1 red and 3 blue pens. You pick a pen, put it back, and pick again. Find the probability that both pens are blue.
Show answer
34 × 34 = 916.
More questions like this
- Fill in the blanks.
- In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).
- Which of the following experiments have equally likely outcomes? Explain.
- Write the sample space and calculate the probability based on the given information.
- A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?