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An experiment consists of selecting a ball from a bag and spinning a coin - Edexcel - A-Level Maths Statistics - Question 2 - 2010 - Paper 2
Question 2
An experiment consists of selecting a ball from a bag and spinning a coin. The bag contains 5 red balls and 7 blue balls. A ball is selected at random from the bag, ... show full transcript
Worked Solution & Example Answer:An experiment consists of selecting a ball from a bag and spinning a coin - Edexcel - A-Level Maths Statistics - Question 2 - 2010 - Paper 2
Step 1
Complete the tree diagram
Answer
For the tree diagram:
- Red Ball Selection:
- Probability = (\frac{5}{12})
- Heads (biassed coin): Probability = (\frac{2}{3})
- Tails (biassed coin): Probability = (\frac{1}{3})
- Blue Ball Selection:
- Probability = (\frac{7}{12})
- Heads (fair coin): Probability = (\frac{1}{2})
- Tails (fair coin): Probability = (\frac{1}{2})
So, the completed tree diagram is as follows:
Ball
|
+--- Red (\(\frac{5}{12}\))
| +--- Heads (\(\frac{2}{3}\))
| | Outcome: Heads
| +--- Tails (\(\frac{1}{3}\))
| Outcome: Tails
|
+--- Blue (\(\frac{7}{12}\))
+--- Heads (\(\frac{1}{2}\))
| Outcome: Heads
+--- Tails (\(\frac{1}{2}\))
Outcome: Tails
The probabilities for the outcomes are:
- Red Heads: (\frac{5}{12} \times \frac{2}{3} = \frac{5}{18})
- Red Tails: (\frac{5}{12} \times \frac{1}{3} = \frac{5}{36})
- Blue Heads: (\frac{7}{12} \times \frac{1}{2} = \frac{7}{24})
- Blue Tails: (\frac{7}{12} \times \frac{1}{2} = \frac{7}{24})
Step 2
Find the probability that Shivani obtains a head
Answer
To find the total probability that Shivani obtains a head, we consider both cases:
-
Probability of Red and Heads: [ P(RH) = P(R) \times P(H|R) = \frac{5}{12} \times \frac{2}{3} = \frac{5}{18} ]
-
Probability of Blue and Heads: [ P(BH) = P(B) \times P(H|B) = \frac{7}{12} \times \frac{1}{2} = \frac{7}{24} ]
Now, adding these probabilities: [ P(H) = P(RH) + P(BH) = \frac{5}{18} + \frac{7}{24} ]
To add these fractions, we need a common denominator: [ \frac{5}{18} = \frac{20}{72} \quad \text{and} \quad \frac{7}{24} = \frac{21}{72} ]
Thus: [ P(H) = \frac{20}{72} + \frac{21}{72} = \frac{41}{72} \approx 0.569 ]
Step 3
Find the probability that Tom selected a red ball
Answer
We use Bayes' theorem for this calculation:
[ P(R|H) = \frac{P(H|R) \times P(R)}{P(H)} ]
Where:
- (P(H|R) = \frac{2}{3})
- (P(R) = \frac{5}{12})
- (P(H) = \frac{41}{72})
Plugging in the values: [ P(R|H) = \frac{\left(\frac{2}{3}\right) \times \left(\frac{5}{12}\right)}{\frac{41}{72}} ] [ =\frac{\frac{10}{36}}{\frac{41}{72}} = \frac{10 \times 72}{36 \times 41} = \frac{720}{1476} \approx \frac{20}{41} \approx 0.488 ]
Step 4
Find the probability that Shivani’s ball colour matches Tom’s
Answer
Since Shivani and Tom both select their balls independently from the same bag:
-
Probability both select Red: [ P(RR) = P(R) \times P(R) = \left(\frac{5}{12}\right) \times \left(\frac{5}{12}\right) = \frac{25}{144} ]
-
Probability both select Blue: [ P(BB) = P(B) \times P(B) = \left(\frac{7}{12}\right) \times \left(\frac{7}{12}\right) = \frac{49}{144} ]
Now, adding these probabilities gives: [ P(S=T) = P(RR) + P(BB) = \frac{25}{144} + \frac{49}{144} = \frac{74}{144} = \frac{37}{72} \approx 0.514 ]