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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
In the space below, draw a tree diagram to illustrate all the possible outcomes and associated probabilities. State your probabilities clearly.
Answer
To illustrate the possible outcomes, we can draw a tree diagram. The first step involves drawing the first draw from the jar, which contains:
- 2 Red beads (R1, R2)
- 1 Blue bead (B)
- 1 Green bead (G)
The probabilities for the first draw are:
- P(R1) = P(R2) = ( \frac{2}{4} = \frac{1}{2} )
- P(B) = ( \frac{1}{4} )
- P(G) = ( \frac{1}{4} )
From each of these first options, second draws are made based on the first draw:
-
If Red (R1 or R2) is drawn first:
- Remaining beads: 1 Red, 1 Blue, 1 Green
- P(Red) = ( \frac{1}{3} )
- P(Blue) = ( \frac{1}{3} )
- P(Green) = ( \frac{1}{3} )
- Remaining beads: 1 Red, 1 Blue, 1 Green
-
If Blue is drawn first:
- Remaining beads: 2 Red, 1 Green
- P(Red) = ( \frac{2}{3} )
- P(Green) = ( \frac{1}{3} )
- Remaining beads: 2 Red, 1 Green
-
If Green is drawn first:
- Remaining beads: 2 Red, 1 Blue
- P(Red) = ( \frac{2}{3} )
- P(Blue) = ( \frac{1}{3} )
- Remaining beads: 2 Red, 1 Blue
The diagram and associated probabilities clearly outline the likelihood of each outcome.
Step 2
Find the probability that a blue bead and a green bead are drawn from the jar.
Answer
To find the probability that a blue bead (B) and a green bead (G) are drawn, we must consider the two possible scenarios:
-
Drawing Blue first and then Green (BG):
- The probability of drawing Blue first is ( P(B) = \frac{1}{4} )
- Then, the probability of drawing Green next is ( P(G|B) = \frac{1}{3} )
- Thus, the combined probability for this sequence is: [ P(BG) = P(B) \times P(G|B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} ]
-
Drawing Green first and then Blue (GB):
- The probability of drawing Green first is ( P(G) = \frac{1}{4} )
- Then, the probability of drawing Blue next is ( P(B|G) = \frac{1}{3} )
- Thus, the combined probability for this sequence is: [ P(GB) = P(G) \times P(B|G) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} ]
Finally, we sum both probabilities to find the total probability of drawing a Blue and a Green bead: [ P(BG \cup GB) = P(BG) + P(GB) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} ]