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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 draw the tree diagram, we start with the first selection:
- First Ball:
- Blue (9 out of 12 total): Probability = ( \frac{9}{12} )
- Red (3 out of 12 total): Probability = ( \frac{3}{12} )
Next, for the second selection, we branch out based on the result of the first ball:
-
If the first ball is Blue:
- Second Ball Blue: Probability = ( \frac{8}{11} )
- Second Ball Red: Probability = ( \frac{3}{11} )
-
If the first ball is Red:
- Second Ball Blue: Probability = ( \frac{9}{11} )
- Second Ball Red: Probability = ( \frac{2}{11} )
The complete tree diagram is illustrated below:
First Selection
/ \
Blue Red
(9/12) (3/12)
/ \ / \
B R B R
(8/11)(3/11)(9/11)(2/11)
Step 2
the second ball selected is red.
Answer
To find the probability that the second ball selected is red, we consider the scenarios where the second ball can be red:
-
The first ball is blue and the second ball is red:
- Probability = ( \frac{9}{12} \times \frac{3}{11} = \frac{27}{132} )
-
The first ball is red and the second ball is red:
- Probability = ( \frac{3}{12} \times \frac{2}{11} = \frac{6}{132} )
Now, we add these probabilities together:
[ P(\text{second ball is red}) = \frac{27}{132} + \frac{6}{132} = \frac{33}{132} = \frac{1}{4} ]
Thus, the probability that the second ball selected is red is ( \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 balls are red given that the second ball is red, we use Bayes' theorem:
[ P(\text{Both are red | Second ball is red}) = \frac{P(\text{Both are red})}{P(\text{Second is red})} ]
From previous calculations:
- ( P(\text{Both are red}) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132} )
- ( P(\text{Second is red}) = \frac{1}{4} = \frac{33}{132} )
Now, substituting into Bayes' theorem:
[ P(\text{Both are red | Second ball is red}) = \frac{\frac{6}{132}}{\frac{33}{132}} = \frac{6}{33} = \frac{2}{11} ]
Thus, the final answer is ( \frac{2}{11} ).