Mathematics, 13.02.2020 22:58, savannahwonder

The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medicare service. For this group, 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012). Suppose 10 first-round appeals have just been received by a Medicare appeal office. a.Compute the probability that none of the appeals will be successful. b.Compute the probability that exactly one of the appeals will be successful. c.What is the probability that at least two of the appeals will be successful?d. What is the probability that more than half of the appeals will be successful?

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answered: danielle4124

Using the binomial distribution, it is found that:

a. 0.006 = 0.06% probability that none of the appeals will be successful.b. 0.0403 = 4.03% probability that exactly one of the appeals will be successful.c. 0.0463 = 4.63% probability that at least two of the appeals will be successful.d. 0.1663 = 16.63% probability that more than half of the appeals will be successful.

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For each appeal, there are only two possible outcomes, either it was accepted, or it was not. The probability of an appeal being accepted is independent of any other appeal, which means that the binomial probability distribution is used to solve this question.

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Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success.

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40% of the appeals are successful, thus p = 0.4

10 appeals, thus, n = 10

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Item a:

This is P(X = 0), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.006$

0.006 = 0.06% probability that none of the appeals will be successful.

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Item b:

This is P(X = 1), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

0.0403 = 4.03% probability that exactly one of the appeals will be successful.

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Item c:

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

P(X < 2) = P(X = 0) + P(X = 1)

Thus

P(X < 2) = P(X = 0) + P(X = 1) = 0.006 + 0.0403 = 0.0463

0.0463 = 4.63% probability that at least two of the appeals will be successful.

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Item d:

This is:

P(X 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.4)^{6}.(0.6)^{4} = 0.1115$

$P(X = 7) = C_{10,7}.(0.4)^{7}.(0.6)^{3} = 0.0425$

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

P(X 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1115 + 0.0425 + 0.0106 + 0.0016 + 0.0001 = 0.1663

0.1663 = 16.63% probability that more than half of the appeals will be successful.

A similar problem is given at link

answered: bloop3r

(a) P (X = 0) = 0.006

(b) P (X = 1) = 0.0403

(d) P (X > 5) = 0.1664

Step-by-step explanation:

Let X = number of successful appeals.

The probability that an appeal is successful is, p = 0.40.

The sample size is, n = 10.

The random variable $X\sim Bin(n=10, p=0.40)$ .

The probability function of a Binomial distribution is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x}$

(a)

The probability that none of the appeals is successful is:

$P(X=0)={10\choose 0}(0.40)^{0}(1-0.40)^{10-0}\\=1\times1\times0.0060466176\\=0.0060466176\\\approx0.006$

Thus, the probability that none of the appeals is successful is 0.006.

(b)

The probability that exactly one of the appeals will be successful is:

$P(X=1)={10\choose 1}(0.40)^{1}(1-0.40)^{10-1}\\=10\times0.40\times0.010077696\\=0.040310784\\\approx0.0403$

Thus, the probability that exactly one of the appeals will be successful is 0.0403.

(c)

The probability that at least two of the appeals will be successful is:

$P(X\geq 2)=1-P(X$

Thus, the probability that at least two of the appeals will be successful is 0.9537.

(d)

The probability that more than half of the appeals will be successful is:

$P(X 5)=1-P(X\leq 5)\\=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)+P(X=5)\\=1-[{10\choose 0}(0.40)^{0}(1-0.40)^{10-0}]-[{10\choose 1}(0.40)^{1}(1-0.40)^{10-1}]\\-[{10\choose 2}(0.40)^{2}(1-0.40)^{10-2}]-[{10\choose 3}(0.40)^{3}(1-0.40)^{10-3}]\\-[{10\choose 4}(0.40)^{4}(1-0.40)^{10-4}]-[{10\choose 5}(0.40)^{5}(1-0.40)^{10-5}]\\=1 - 0.006-0.0403-0.1209-0.2150-0.2508-0.2006\\=0.1664$