g the center for medicare services reported that there were 295,000 reported appeals for hospitalization and other medicare part a services. for this group, 40% of the first round appeals were successful. suppose ten first round appeals were just received. the results follow a binomial distribution. what is the probability of getting more than two heads? enter answer to four decimal places.

There are just two options for each appeal: either it has been accepted or it wasn't.
A binomial probability is employed to answer this question since the chances that one appeal will be permitted is independent of the possibility that any other appeals will be granted.

The probability of precisely x successes per n repeated trials is known as a binomial probability, and X has only two possible outcomes.

P(X = x) = Cn,ₓ.pˣ(1 - p)ⁿ⁻ˣ ...eq 1

Where Cn,ₓ represents the number of distinct combinations of x items drawn from a group of n elements, as determined by the formula below.

Cn,ₓ = n!/x!(n - x)!

Where p is the likelihood that anything will happen.

10 appeals are necessary; n = 10.

Because 40% of appeals are successful; p = 0.4.

For the probability of getting at least two heads.

P(X ≥ 2) = 1 - P(X < 1)

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

Put the values in eq 1 and get the values for X.

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

Now,

P(X ≥ 2) = 1 - 0.0463

P(X ≥ 2) = 0.9537

Thus, the probability that of getting more than two heads is 0.9537.

To know more about the binomial probability, here

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