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A jar contains 2 red, 1 blue and 1 green bead - Edexcel - A-Level Maths Statistics - Question 1 - 2010 - Paper 1
Question 1
A jar contains 2 red, 1 blue and 1 green bead. Two beads are drawn at random from the jar without replacement. (a) In the space below, draw a tree diagram to illust... show full transcript
Worked Solution & Example Answer:A jar contains 2 red, 1 blue and 1 green bead - Edexcel - A-Level Maths Statistics - Question 1 - 2010 - Paper 1
Step 1
Draw the Tree Diagram
Answer
To illustrate the possible outcomes, we start by drawing a tree diagram. The first branching represents the selection of the first bead:
-
First bead can be:
- Red (with probability rac{2}{4})
- Blue (with probability rac{1}{4})
- Green (with probability rac{1}{4})
-
For each of these outcomes, draw branches for the second bead selection:
- From a Red bead:
- Red (with probability rac{1}{3})
- Blue (with probability rac{1}{3})
- Green (with probability rac{1}{3})
- From a Blue bead:
- Red (with probability rac{2}{3})
- Green (with probability rac{1}{3})
- From a Green bead:
- Red (with probability rac{2}{3})
- Blue (with probability rac{1}{3})
- From a Red bead:
Thus, the full tree diagram will be:
First Bead
├─ Red (2/4)
│ ├─ Red (1/3)
│ ├─ Blue (1/3)
│ └─ Green (1/3)
├─ Blue (1/4)
│ ├─ Red (2/3)
│ └─ Green (1/3)
└─ Green (1/4)
├─ Red (2/3)
└─ Blue (1/3)
Step 2
Find the probability that a blue bead and a green bead are drawn
Answer
To calculate this probability, we consider the two cases where a blue bead and a green bead can be drawn:
-
Case 1 (Blue first, then Green): Probability = Probability(Blue) * Probability(Green given Blue)
= rac{1}{4} * rac{1}{3} = rac{1}{12} -
Case 2 (Green first, then Blue): Probability = Probability(Green) * Probability(Blue given Green)
= rac{1}{4} * rac{2}{3} = rac{1}{6}
Now, the total probability of drawing a blue and a green bead (regardless of order) is:
P(Blue ext{ and } Green) = rac{1}{12} + rac{1}{6} = rac{1}{12} + rac{2}{12} = rac{3}{12} = rac{1}{4}