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A bag contains 9 blue balls and 3 red balls - Edexcel - A-Level Maths Statistics - Question 4 - 2006 - Paper 1
Question 4
A bag contains 9 blue balls and 3 red balls. A ball is selected at random from the bag and its colour is recorded. The ball is not replaced. A second ball is selecte... show full transcript
Worked Solution & Example Answer:A bag contains 9 blue balls and 3 red balls - Edexcel - A-Level Maths Statistics - Question 4 - 2006 - Paper 1
Step 1
Draw a tree diagram to represent the information.
Answer
To represent the information, we construct a tree diagram.
- Start with a single point representing the selection of the first ball.
- From this point, branch out to two outcomes: selecting a blue ball or selecting a red ball.
- Probability of selecting a blue ball: ( P(Blue) = \frac{9}{12} = \frac{3}{4} )
- Probability of selecting a red ball: ( P(Red) = \frac{3}{12} = \frac{1}{4} )
- From each of these outcomes, draw branches for the selection of the second ball, noting that the total number of balls now changes:
- If the first ball was blue (9 blue - 1 = 8 blue, still 3 red):
- Probability of selecting blue second: ( P(Blue|First Blue) = \frac{8}{11} )
- Probability of selecting red second: ( P(Red|First Blue) = \frac{3}{11} )
- If the first ball was red (3 red - 1 = 2 red, still 9 blue):
- Probability of selecting blue second: ( P(Blue|First Red) = \frac{9}{11} )
- Probability of selecting red second: ( P(Red|First Red) = \frac{2}{11} )
- If the first ball was blue (9 blue - 1 = 8 blue, still 3 red):
- The completed tree shows all possible outcomes with their probabilities.
Step 2
the second ball selected is red.
Answer
To find the probability that the second ball selected is red, we consider both paths that lead to selecting a red ball second:
-
If the first ball is blue:
- Probability of first selecting blue and then red is: [ P(Blue, Red) = P(Blue) \times P(Red|First Blue) = \frac{9}{12} \times \frac{3}{11} = \frac{27}{132} ]
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If the first ball is red:
- Probability of first selecting red and then red is: [ P(Red, Red) = P(Red) \times P(Red|First Red) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132} ]
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Now, sum both probabilities for the event that the second ball is red: [ P(Second\ is\ Red) = P(Blue, Red) + P(Red, Red) = \frac{27}{132} + \frac{6}{132} = \frac{33}{132} = \frac{1}{4} ]
Step 3
both balls selected are red, given that the second ball selected is red.
Answer
To find the conditional probability that both drawn balls are red given that the second ball is red, we use:
[ P(Both\ Red | Second\ Red) = \frac{P(Both\ Red)}{P(Second\ Red)} ]
- We already calculated ( P(Second\ is\ Red) = \frac{1}{4} ).
- Now, find ( P(Both\ Red) ): [ P(Both\ Red) = P(Red) \times P(Red|First\ Red) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132} = \frac{1}{22} ]
- Therefore, substituting into our original equation gives: [ P(Both\ Red | Second\ Red) = \frac{\frac{1}{22}}{\frac{1}{4}} = \frac{1}{22} \times \frac{4}{1} = \frac{4}{22} = \frac{2}{11} ]