(a) To find the probability that all 10 trustworthy individuals pass the polygraph test,
we can use the binomial probability formula:
P(X = 10) = C(10, 10) * (0.15)^10 * (1 - 0.15)^(10 - 10)
Calculating the values:
C(10, 10) = 1 (since choosing all 10 out of 10 is only one possibility)
(0.15)^10 ≈ 0.0000000778
(1 - 0.15)^(10 - 10) = 1 (anything raised to the power of 0 is 1)
P(X = 10) ≈ 1 * 0.0000000778 * 1 ≈ 0.0000000778
The probability that all 10 trustworthy individuals pass the polygraph test is approximately 0.0000000778.
(b) To find the probability that more than 2 trustworthy individuals fail the test, we need to calculate the probability of exactly 0, 1, and 2 individuals failing and subtract it from 1 (to find the complementary probability).
P(more than 2 fail, even though all are trustworthy) = 1 - P(X = 0) - P(X = 1) - P(X = 2)
Using the binomial probability formula:
P(X = 0) = C(10, 0) * (0.15)^0 * (1 - 0.15)^(10 - 0)
P(X = 1) = C(10, 1) * (0.15)^1 * (1 - 0.15)^(10 - 1)
P(X = 2) = C(10, 2) * (0.15)^2 * (1 - 0.15)^(10 - 2)
Calculating the values:
C(10, 0) = 1
C(10, 1) = 10
C(10, 2) = 45
(0.15)^0 = 1
(0.15)^1 = 0.15
(0.15)^2 ≈ 0.0225
(1 - 0.15)^(10 - 0) = 0.85^10 ≈ 0.1967
(1 - 0.15)^(10 - 1) = 0.85^9 ≈ 0.2209
(1 - 0.15)^(10 - 2) = 0.85^8 ≈ 0.2476
P(more than 2 fail, even though all are trustworthy) = 1 - 1 * 0.1967 - 10 * 0.15 * 0.2209 - 45 * 0.0225 * 0.2476 ≈ 0.0004
The probability that more than 2 trustworthy individuals fail the polygraph test, even though all are trustworthy, is approximately 0.0004.
(c) The mean (expected value) of a binomial distribution is given by μ = np, where n is the number of trials (400 agents tested) and p is the probability of success (the probability of failing for a trustworthy agent, which is 0.15).
Mean = μ = np = 400 * 0.15 = 60
The standard deviation of a binomial distribution is given by σ = sqrt(np(1-p)).
Standard deviation = σ = sqrt(400 * 0.15 * (1 - 0.15)) ≈ 4
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Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2
The general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
Given differential equations are:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x²
To find the general solution to the given differential equations, we will solve these equations one by one.
(i) 16y'' - 8y' + y = 0
The characteristic equation is:
16m² - 8m + 1 = 0
Solving this quadratic equation, we get m = 1/4, 1/4
Hence, the general solution of the given differential equation is:
y = c₁e^(x/4) + c₂xe^(x/4)..................................................(1)
(ii) y" + y' - 2y = 0
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2
Hence, the general solution of the given differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(2)
(iii) y" + y' - 2y = x²
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2.
The complementary function (CF) of this differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(3)
Now, we will find the particular integral (PI). Let's assume that the PI of the differential equation is of the form:
y = Ax² + Bx + C
Substituting the value of y in the given differential equation, we get:
2A - 4A + 2Ax² + 4Ax - 2Ax² = x²
Equating the coefficients of x², x, and the constant terms on both sides, we get:
2A - 2A = 1,
4A - 4A = 0, and
2A = 0
Solving these equations, we get
A = 1/2,
B = 0, and
C = 0
Hence, the particular integral of the given differential equation is:
y = (1/2)x²..................................................(4)
The general solution of the given differential equation is the sum of CF and PI.
Hence, the general solution is:
y = c₁e^x + c₂e^(-2x) + (1/2)x²..................................................(5)
Conclusion: Therefore, the general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
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The particular solution is: y = -1/2 x². The general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
The general solution of the given differential equations are:
Given differential equation: 16y'' - 8y' + y = 0
The auxiliary equation is: 16m² - 8m + 1 = 0
On solving the above quadratic equation, we get:
m = 1/4, 1/4
∴ General solution of the given differential equation is:
y = c1 e^(x/4) + c2 x e^(x/4)
Given differential equation: y" + y' - 2y = 0
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:
m = -2, 1
∴ General solution of the given differential equation is:
y = c1 e^(-2x) + c2 e^(x)
Given differential equation: y" + y' - 2y = x²
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:m = -2, 1
∴ The complementary solution is:y = c1 e^(-2x) + c2 e^(x)
Now we have to find the particular solution, let us assume the particular solution of the given differential equation:
y = ax² + bx + c
We will use the method of undetermined coefficients.
Substituting y in the differential equation:y" + y' - 2y = x²a(2) + 2a + b - 2ax² - 2bx - 2c = x²
Comparing the coefficients of x² on both sides, we get:-2a = 1
∴ a = -1/2
Comparing the coefficients of x on both sides, we get:-2b = 0 ∴ b = 0
Comparing the constant terms on both sides, we get:2c = 0 ∴ c = 0
Thus, the particular solution is: y = -1/2 x²
Now, the general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
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