Question

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?

Answer 1

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 = 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:

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

Thus, the probability that more than half of the appeals will be successful is 0.1664.