A water well is to be drilled in the desert where the soil is either rock, clay or sand. The probability of rock P(R)equals=0.
53. The clay probability is P(C)equals=0.21. The sand probability is P(S)equals=0.26. If the soil is rock, a geological test gives a positive result with 35% accuracy. If it is clay, this test gives a positive result with 48% accuracy. The test gives a 75% accuracy for sand.
Given the test is positive, what is the probability that the soil is clay, P(clay | positive)? Use Bayes' rule to find the indicated probability.
Given the test is positive, what is the probability that the soil is clay, P(clay | positive)? Use Bayes' rule to find the indicated probability.
1 answer:
The tree diagram for the probability is shown below
P(Clay|Positive) is read 'Probability of Clay given the result is Positive'.
This is a case of conditional probability.
The formula for conditional probability is given as
P(Clay|Positive) = P(Clay∩Positive) ÷ P(Positive)
P(Clay∩Positive) = 0.21×0.48 = 0.1008
P(Positive) = P(Rock∩Positive) + P(Clay∩Positive) + P(Sand∩Positive)
P(Positive) = (0.53×0.53) + (0.21×0.48) + (0.26×0.75)
P(Positive) = 0.2809 + 0.1008 + 0.195
P(Positive) = 0.5767
Hence,
P(Clay|Positive) = 0.1008÷0.5767 = 0.175 (rounded to 3 decimal place)
P(Clay|Positive) is read 'Probability of Clay given the result is Positive'.
This is a case of conditional probability.
The formula for conditional probability is given as
P(Clay|Positive) = P(Clay∩Positive) ÷ P(Positive)
P(Clay∩Positive) = 0.21×0.48 = 0.1008
P(Positive) = P(Rock∩Positive) + P(Clay∩Positive) + P(Sand∩Positive)
P(Positive) = (0.53×0.53) + (0.21×0.48) + (0.26×0.75)
P(Positive) = 0.2809 + 0.1008 + 0.195
P(Positive) = 0.5767
Hence,
P(Clay|Positive) = 0.1008÷0.5767 = 0.175 (rounded to 3 decimal place)
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Therefore, parabola (1) is wider than parabola (2).
Part B:
A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.
Given parabola (1) to be, and parabola (2) to be
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Part C:
For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).
Given parabola (1) to be, and parabola (2) to be
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Therefore, the parabola that would be furthest left on the x-axis is parabola (1).
Part D:
For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).
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Therefore, the parabola that would be highest on the y-axis is parabola (2).
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Therefore, parabola (1) is wider than parabola (2).
Part B:
A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.
Given parabola (1) to be
Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).
Part C:
For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).
Given parabola (1) to be
Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.
Therefore, the parabola that would be furthest left on the x-axis is parabola (1).
Part D:
For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).
Given parabola (1) to be
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Therefore, the parabola that would be highest on the y-axis is parabola (2).