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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 below to show the possible outcomes and associated probabilities.
Answer
To complete the tree diagram, we first determine the total number of balls in the bag. The total is 5 red balls + 7 blue balls = 12 balls.
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For the red ball branch:
- Probability of selecting a red ball = \frac{5}{12}
- The biased coin has a probability of landing heads = \frac{2}{3}
- Probability of obtaining heads after a red ball = \frac{5}{12} \times \frac{2}{3} = \frac{10}{36} = \frac{5}{18}
- Probability of obtaining tails after a red ball = 1 - \frac{2}{3} = \frac{1}{3}
- Probability of obtaining tails after a red ball = \frac{5}{12} \times \frac{1}{3} = \frac{5}{36}
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For the blue ball branch:
- Probability of selecting a blue ball = \frac{7}{12}
- The fair coin has a probability of landing heads = \frac{1}{2}
- Probability of obtaining heads after a blue ball = \frac{7}{12} \times \frac{1}{2} = \frac{7}{24}
- Probability of obtaining tails after a blue ball = 1 - \frac{1}{2} = \frac{1}{2}
- Probability of obtaining tails after a blue ball = \frac{7}{12} \times \frac{1}{2} = \frac{7}{24}
Thus the completed tree diagram should show:
- Red: Heads (\frac{5}{18}) and Tails (\frac{5}{36})
- Blue: Heads (\frac{7}{24}) and Tails (\frac{7}{24})
Step 2
Find the probability that she obtains a head.
Answer
The probability of obtaining a head can be calculated by summing the probabilities of obtaining heads from both branches:
- From the red branch: ( P(RH) = \frac{5}{18} )
- From the blue branch: ( P(BH) = \frac{7}{24} )
To add these probabilities, we need a common denominator. The least common multiple of 18 and 24 is 72:
- Convert ( P(RH) ): ( \frac{5}{18} = \frac{20}{72} )
- Convert ( P(BH) ): ( \frac{7}{24} = \frac{21}{72} )
Now we can find:
( P(H) = P(RH) + P(BH) = \frac{20}{72} + \frac{21}{72} = \frac{41}{72} )
Thus, the probability that Shivani obtains a head is ( \frac{41}{72} )
Step 3
Given that Tom selected a ball at random and obtained a head when he spun the appropriate coin, find the probability that Tom selected a red ball.
Answer
Let ( R ) be the event that Tom selects a red ball and ( H ) be the event that he obtains a head. We are looking for ( P(R|H) ), which can be found using Bayes' theorem:
[ P(R|H) = \frac{P(H|R) \cdot P(R)}{P(H)} ]
- We already found ( P(H) = \frac{41}{72} ).
- From previous calculations:
- ( P(H|R) = \frac{2}{3} )
- ( P(R) = \frac{5}{12} )
Substituting these values into Bayes' theorem:
[ P(R|H) = \frac{ \frac{2}{3} \cdot \frac{5}{12} }{ \frac{41}{72} } = \frac{ \frac{10}{36} }{ \frac{41}{72} } = \frac{10 \cdot 72}{36 \cdot 41} = \frac{20}{41} ]
Thus, the probability that Tom selected a red ball given that he obtained a head is ( \frac{20}{41} ).
Step 4
Find the probability that the colour of the ball Shivani selects is the same as the colour of the ball Tom selects.
Answer
To find the probability that both Shivani and Tom select the same colour ball, we consider the two scenarios:
- Both select red balls:
- Probability that Shivani selects a red ball = ( P(R) = \frac{5}{12} )
- Probability that Tom also selects a red ball = ( P(R) = \frac{5}{12} )
Thus, probability for both selecting red: ( P(R ext{ and } R) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144} )
- Both select blue balls:
- Probability that Shivani selects a blue ball = ( P(B) = \frac{7}{12} )
- Probability that Tom also selects a blue ball = ( P(B) = \frac{7}{12} )
Thus, probability for both selecting blue: ( P(B ext{ and } B) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144} )
Now, we can sum these probabilities:
[ P( ext{same colour}) = P(R ext{ and } R) + P(B ext{ and } B) = \frac{25}{144} + \frac{49}{144} = \frac{74}{144} = \frac{37}{72} ]
Therefore, the probability that the colour of the ball Shivani selects is the same as the colour of the ball Tom selects is ( \frac{37}{72} ).