A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.
- (i) What is the probability of drawing a red ball and then a blue ball?
- (ii) What is the probability of drawing 2 blue balls?
Step-by-step solution
Idea: Because the first ball is laid aside, the second draw has only 8 balls, and its probabilities depend on what came first. Multiply the probabilities along a path to get the probability of that outcome.
(i) What is the probability of drawing a red ball and then a blue ball?
- Tree, first draw (9 balls): Red 49, Blue 59.½ mark
- Second draw (8 balls left):
after Red (3 red, 5 blue left): Red 38, Blue 58;
after Blue (4 red, 4 blue left): Red 48, Blue 48.1 mark - Outcomes: RR, RB, BR, BB with probabilities 1272, 2072, 2072, 2072 (multiply along each path).½ mark
- Red then blue is the path R → B: P = 49 × 58 = 2072 = 518.
(Counting check: label the balls; there are 9 × 8 = 72 equally likely ordered draws, and 4 × 5 = 20 of them are red then blue.)1 mark
(ii) What is the probability of drawing 2 blue balls?
- Path B → B: P = 59 × 48 = 2072 = 518. (Counting: 5 × 4 = 20 of the 72 ordered draws.)1 mark
Check: All four paths add to 1: (12 + 20 + 20 + 20)/72 = 7272 ✓.
Answer to write in the exam
(i)
1st draw: P(R) = 49, P(B) = 59
2nd draw after R: P(R) = 38, P(B) = 58; after B: P(R) = 48, P(B) = 48
P(RR) = 1272, P(RB) = 2072, P(BR) = 2072, P(BB) = 2072
P(red then blue) = 49 × 58 = 2072
∴ P(red then blue) = 518
(ii)
P(BB) = 59 × 48 = 2072
∴ P(2 blue balls) = 518
Common mistakes that cost marks
- Using 59 again for the second draw. The first ball is laid aside, so only 8 balls are left and the second-draw fractions have denominator 8.
- Forgetting that after a blue ball, only 4 blue balls are left (48, not 58).
- In (i), adding the red-then-blue and blue-then-red paths. The question asks for red then blue, which is only one path.
How this can come in the exam
A box has 3 white and 2 black counters. Two counters are drawn one after the other without replacement. The probability that both are black is
- 110
- 425
- 15
- 25
Show answer
(A) 110
25 × 14 = 220 = 110.
A bag has 3 green and 2 yellow balls. One ball is drawn and not replaced, then another is drawn. Draw a tree diagram and find the probability that the two balls are of different colours.
Show answer
1st: G 35, Y 25. 2nd after G: G 24, Y 24; after Y: G 34, Y 14. P(GY) = 35 × 24 = 620; P(YG) = 25 × 34 = 620. P(different) = 1220 = 35.Try one yourself
With the same basket (4 red, 5 blue, no replacement), find the probability of drawing 2 red balls.
Show answer
49 × 38 = 1272 = 16.
More questions like this
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- Write the sample space and calculate the probability based on the given information.
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- List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.
- Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?