The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic

Question

2 months ago

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The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic

are 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?

Mathematics

1 answer:

lawyer [12.5K] · Answer 1 · 2026-02-17T03:50:57+03:00

Answer:

(a) P (X = 0) = 0.006

(b) P (X = 1) = 0.0403

(c) P (X ≥ 2) = 0.9537

(d) P (X > 5) = 0.1664

Step-by-step explanation:

Let X be the number of successful appeals.

The probability of a successful appeal is p = 0.40.

The sample size is n = 10.

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

The probability function of a Binomial distribution is represented by:

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

(a)

The likelihood that none of the appeals succeeds is:

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

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

(b)

The likelihood that exactly one appeal succeeds 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 of a single successful appeal is 0.0403.

(c)

The probability of at least two appeals being successful is:

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

So, the probability that at least two appeals are successful is 0.9537.

(d)

The probability of more than half of the appeals being successful is:

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

Thus, the probability that more than half succeed is 0.1664.