3. Magnetic resonance imaging (MRI) is a process that produces internal body images using a strong magnetic field. Some patients become claustrophobic and require sedation because they are required to lie within a small, enclosed space during the MRI test. Suppose 29% of all patients undergoing MRI testing require sedation due to claustrophobia. If five patients are selected at random, using the binomial probability distribution formula, find the probability that the number of patients in these five who require sedation is

a) Exactly 2
b) None
c) Exactly 4

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To find the probabilities, we will use the binomial probability distribution formula:

P(X=k) = nCk * p^k * (1-p)^(n-k)

Where:
P(X=k) is the probability of having exactly k successes,
n is the number of trials,
k is the number of successful outcomes,
p is the probability of success in a single trial,
(1-p) is the probability of failure in a single trial,
nCk is the number of combinations of n items taken k at a time.

a) Exactly 2 patients who require sedation:
In this case, k = 2, n= 5, p = 0.29

P(X=2) = 5C2 * 0.29^2 * (1-0.29)^(5-2)
= 10 * 0.29^2 * 0.71^3
≈ 0.30

Therefore, the probability that exactly 2 patients require sedation is approximately 0.30.

b) None of the patients require sedation:
In this case, k = 0, n = 5, p = 0.29

P(X=0) = 5C0 * 0.29^0 * (1-0.29)^(5-0)
= 1 * 1 * 0.71^5
≈ 0.29^0 * 0.71^5
≈ 0.71^5
≈ 0.168

Therefore, the probability that none of the patients require sedation is approximately 0.168.

c) Exactly 4 patients who require sedation:
In this case, k = 4, n= 5, p = 0.29

P(X=4) = 5C4 * 0.29^4 * (1-0.29)^(5-4)
= 5 * 0.29^4 * 0.71^1
≈ 0.012

Therefore, the probability that exactly 4 patients require sedation is approximately 0.012.

To find the probability using the binomial probability distribution formula, we need to know the number of trials (n), the probability of success (p), and the number of successes (x) that we're interested in.

In this case, the number of trials (n) is 5, since we are selecting five patients at random. The probability of each patient requiring sedation (p) is given as 0.29, or 29%.

a) To find the probability that exactly 2 patients require sedation, we use the binomial probability distribution formula:

P(X = x) = nCx * p^x * (1 - p)^(n - x)

Where nCx is the number of combinations of n items taken x at a time. So, for this case:
P(X = 2) = 5C2 * (0.29)^2 * (1 - 0.29)^(5 - 2)

To calculate 5C2, we use the combination formula:
5C2 = 5! / (2! * (5-2)!)

Substituting these values into the formula, we get:
P(X = 2) = (5! / (2! * (5-2)!)) * (0.29)^2 * (1 - 0.29)^(5 - 2)

You can calculate this to get the probability that exactly 2 patients require sedation.

b) To find the probability that none of the patients require sedation, we use the same formula:

P(X = x) = nCx * p^x * (1 - p)^(n - x)

In this case, we want to find P(X = 0):

P(X = 0) = 5C0 * (0.29)^0 * (1 - 0.29)^(5 - 0)

Using the combination formula, we get:
P(X = 0) = (5! / (0! * (5-0)!)) * (0.29)^0 * (1 - 0.29)^(5 - 0)

Calculate this to find the probability that none of the patients require sedation.

c) To find the probability that exactly 4 patients require sedation, we again use the same formula:

P(X = x) = nCx * p^x * (1 - p)^(n - x)

In this case, we're interested in P(X = 4):

P(X = 4) = 5C4 * (0.29)^4 * (1 - 0.29)^(5 - 4)

Using the combination formula, we get:
P(X = 4) = (5! / (4! * (5-4)!)) * (0.29)^4 * (1 - 0.29)^(5 - 4)

Calculate this to find the probability that exactly 4 patients require sedation.