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Tju [1.3M]
3 years ago
13

Three candidates run for different offices in different cities. Each has a one in three chance of being elected in his/her city.

What is the probability that at least one of them will be elected?
Mathematics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

P(1) = 1 - 8/27 = 19/27

The probability that at least one of them will be elected is 19/27

Step-by-step explanation:

the probability that at least one of them will be elected = 1 - probability that none of them will be elected.

P(1) = 1 - P(None) .....1

Let P(A), P(B) and P(C) represent the probability for each of the three candidates to be elected .

P(A) = P(B) = P(C) = 1/3

The probability for each of the three candidates not to be elected is

P(A)' = P(B)' = P(C)' = 1 - 1/3 = 2/3

P(None) = P(A)' × P(B)' × P(C)'= 2/3 × 2/3 × 2/3 = 8/27

From equation 1. Substituting the value of P(None)

P(1) = 1 - 8/27 = 19/27

The probability that at least one of them will be elected is 19/27

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Answer:

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Step-by-step explanation:

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To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

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An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

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                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

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Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

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Mean Square Error for the estimator = 6.6311

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Answer:

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Given that we have;

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When x = 8, the point of reflection, the height, f(x) is given as follows;

f(x) = \dfrac{9}{8} \times \left | 8 - 8\right | = 0

When x = 7, the point of reflection, the height, f(x) is given as follows;

f(x) = \dfrac{9}{8} \times \left | 7 - 8\right | = \dfrac{9}{8}

Therefore, given that the point of reflection is at an elevation of 0 relative to the 9 feet of the laser source (pointer), by tan rule, we have;

tan(\theta) = \dfrac{Opposite \ side}{Adjacent \ side} =\dfrac{9}{8} = \dfrac{h}{7}

Where;

h = The height at which the laser meets the wall

h = \dfrac{9 \times  7}{8} =7.875

Given that the wall the laser meets is at the point x with elevation 9/8, the height, y, at which the laser meets the wall is therefore;

y = \dfrac{9 \times  7}{8} - \dfrac{9}{8} = \dfrac{54}{8} = 6.75 \ feet

The height (in feet) at which the laser will impact the wall = 6.75 feet.

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