A spinner is designed so that the score S is given by the following probability distribution - Edexcel - A-Level Maths Statistics - Question 8 - 2011 - Paper 2
Question 8
A spinner is designed so that the score S is given by the following probability distribution.
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Worked Solution & Example Answer:A spinner is designed so that the score S is given by the following probability distribution - Edexcel - A-Level Maths Statistics - Question 8 - 2011 - Paper 2
Step 1
Find the value of p.
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Answer
To find the value of p, we use the property that the sum of probabilities must equal 1:
p+0.25+0.20+0.20+0.20=1
This can be simplified to:
p+0.85=1
Thus, we have:
p=1−0.85=0.15
Step 2
Find E(S).
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Answer
The variance Var(S) can be calculated using:
Var(S)=E(S2)−(E(S))2
Since we have already found that E(S) = 2.45 and E(S²) = 9.45, we substitute these values:
Var(S)=9.45−(2.45)2
Calculating yields:
Var(S)=9.45−6.0025=3.4475.
Step 5
Find the probability that Jess wins after 2 spins.
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Answer
To find the probability that Jess wins after 2 spins, we consider the possible outcomes:
On both spins, if at least one is odd, Jess wins.
The calculations for this would involve finding the total probability of odd scores across the two spins. Let P(odd) be the sum of probabilities of getting an odd score:
P(odd)=p+0.25=0.15+0.25=0.40.
The probability that Jess wins after 2 spins is:
P(Jessextwins)=1−P(Tomextwins)=1−P(oddextonboth)=1−(0.40)2=1−0.16=0.84.
Step 6
Find the probability that Tom wins after exactly 3 spins.
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Answer
To find the probability that Tom wins after exactly 3 spins, we first calculate the total probability that Tom wins on the third spin. This involves considering that Tom must not win on the first two spins but must win on the third. This requires detailed calculations for the probabilities involved:
Count cases of odd/even scores across the spins.
Find the probability that Jess wins after exactly 3 spins.
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For Jess to win after exactly 3 spins, she must be odd for the winning side on the third spin specifically while not winning in the first two spins. This gives:
P(Jessextwinsafter3)=(P(odd)2)(P(Jessextwinonthird))=(0.402)imes(0.40)=0.064.