The national vaccine information center estimates that 90% of americans have had chickenpox by the time they reach adulthood.50 (a) is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled american adults had chickenpox during childhood. (b) calculate the probability that exactly 97 out of 100 randomly sampled american adults had chickenpox during childhood. (c) what is the probability that exactly 3 out of a new sample of 100 american adults have not had chickenpox in their childhood? (d) what is the probability that at least 1 out of 10 randomly sampled american adults have had chickenpox? (e) what is the probability that at most 3 out of 10 randomly sampled american adults have not had chickenpox?

The correct answers are:

A) yes; B) 0.0059; C) 0.0059; D) 1; E) 0.9872.

Explanation:

A) A binomial experiment is one in which the experiment consists of identical trials; each trial results in one of two outcomes, called success and failure; the probability of success remains the same from trial to trial; and the trials are independent.

All of these criteria fit this experiment.

B) The formula for the probability of a binomial experiment is:

[tex] _nC_r\times(p^r)(1-p)^{n-r} [/tex]

where n is the number of trials, r is the number of successes, and p is the probability of success.

In this problem, p = 0.9.

For part B, n = 100 and r = 97:

[tex] _{100}C_{97}(0.9)^{97}(1-0.9)^3
\\=\frac{100!}{97!3!}\times (0.9)^{97}(0.1)^3
\\
\\=161700(0.9)^{0.97}(0.1)^3=0.00589\approx 0.0059 [/tex]

C) We are changing the probability of success this time. Since 90% of people have had chicken pox, then 100%-90% = 1-0.9 = 0.1 have not had chicken pox. For part C, n = 100, r = 3, and p = 0.1:

[tex] _{100}C_3(0.1)^3(1-0.1)^{100-3}
\\
\\=_{100}C_3(0.1)^3(0.9)^{97}
\\=\frac{100!}{97!3!}\times (0.1)^3(0.9)^{97}
\\
\\=161700(0.1)^3(0.9)^{97}=0.00589\approx 0.0059 [/tex]

D) For this part, we want to know the probability that at least 1 person has contracted chicken pox. For this part, p = 0.9, n = 10 and r = 0. We will then subtract this from 1; this will first give us the probability that none of the 10 contracted chicken pox, then subtracting from 1 means that 1 or more people did:

[tex] 1-(_{10}C_0(0.9)^0(1-0.9)^{10-0})
\\
\\=1-(\frac{10!}{0!10!}\times (0.9)^0(0.1)^{10})
\\
\\=1-(1\times 1\times (0.1)^{10})= 1-0 = 1 [/tex]

E) For this part, we find the probability that 3 people, 2 people, 1 person and 0 people have not had chicken pox. The probability p = 0.1; n = 10; and r = 3, 2, 1 and 0, respectively:

[tex] _{10}C_3(0.1)^3(1-0.1)^{10-3}+_{10}C_2(0.1)^2(1-0.1)^{10-2}+
_{10}C_1(0.1)^1(1-0.1)^{10-1}+_{10}C_0(0.1)^0(1-0.1)^{10-0}
\\
\\=_{10}C_3(0.1)^3(0.9)^7+_{10}C_2(0.1)^2(0.9)^8+_{10}C_1(0.1)^1(0.9)^9+
_{10}C_0(0.1)^1(0.9)^{10}
\\
\\120(0.1)^3(0.9)^7+45(0.1)^2(0.9)^8+10(0.1)^1(0.9)^9+1(0.1)^0(0.9)^{10}
\\
\\0.057395628+0.1937102445+0.387420489+0.3486784401
\\
\\=0.9872 [/tex]

The binomial distribution is appropriate for calculating the probability of having a specific number of American adults who had chickenpox during childhood. The probability of exactly 97 out of 100 adults having chickenpox can be calculated using the binomial probability formula. The probability that at least 1 out of 10 adults have had chickenpox and at most 3 out of 10 adults have not had chickenpox can also be calculated using the binomial probability formula.

(a) To determine if the use of the binomial distribution is appropriate, we need to check if the conditions for using it are satisfied: (1) There are only two possible outcomes - having or not having chickenpox. (2) Each trial is independent - one person's chickenpox status does not affect another person's. (3) The probability of having chickenpox is the same for each person. The given information satisfies these conditions, so the binomial distribution is appropriate.

(b) The probability of exactly 97 out of 100 randomly sampled American adults having chickenpox during childhood can be calculated using the binomial probability formula:

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

Where:

P(X = k) is the probability of getting exactly k successes (97 in this case)

C(n, k) is the number of ways to choose k successes out of n trials (100 in this case)

p is the probability of success (probability of having chickenpox = 0.90)

n is the total number of trials (100 in this case)

Using these values, we can calculate:

P(X = 97) = C(100, 97) * 0.90^97 * 0.10^3

= 100 * (0.90)^97 * (0.10)^3

≈ 0.0975

So, the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood is approximately 0.0975 or 9.75%.

(c) The probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood can be calculated using the binomial probability formula:

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

Where:

P(X = k) is the probability of getting exactly k successes (3 in this case)

C(n, k) is the number of ways to choose k successes out of n trials (100 in this case)

p is the probability of success (probability of not having chickenpox = 0.10)

n is the total number of trials (100 in this case)

Using these values, we can calculate:

P(X = 3) = C(100, 3) * 0.10^3 * 0.90^97

= 161,700 * (0.10)^3 * (0.90)^97

≈ 0.0315

So, the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood is approximately 0.0315 or 3.15%.

(d) To calculate the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox, we can use the complement rule: P(at least 1) = 1 - P(none)

Where P(none) is the probability of none of the 10 sampled adults having chickenpox.

Using the binomial formula:

P(X = 0) = C(n, k) * p^k * (1-p)^(n-k)

Where:

P(X = 0) is the probability of getting exactly 0 successes

C(n, k) is the number of ways to choose 0 successes out of n trials (10 in this case)

p is the probability of success (probability of having chickenpox = 0.90)

n is the total number of trials (10 in this case)

Using these values, we can calculate:

P(X = 0) = C(10, 0) * 0.90^0 * 0.10^10

= 1 * (0.90)^0 * (0.10)^10

≈ 0.3487

So, P(none) ≈ 0.3487

Therefore, P(at least 1) = 1 - P(none) = 1 - 0.3487 = 0.6513

So, the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox is approximately 0.6513 or 65.13%.

(e) To calculate the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox, we can add up the probabilities of getting 0, 1, 2, and 3 successes:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

We can use the binomial probability formula to calculate each individual probability:

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

Where:

P(X = k) is the probability of getting exactly k successes (0, 1, 2, or 3 in this case)

C(n, k) is the number of ways to choose k successes out of n trials (10 in this case)

p is the probability of success (probability of not having chickenpox = 0.10)

n is the total number of trials (10 in this case)

Using these values, we can calculate each individual probability:

P(X = 0) = C(10, 0) * 0.10^0 * 0.90^10

P(X = 1) = C(10, 1) * 0.10^1 * 0.90^9

P(X = 2) = C(10, 2) * 0.10^2 * 0.90^8

P(X = 3) = C(10, 3) * 0.10^3 * 0.90^7

Adding up these probabilities, we get:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

≈ 0.9873

So, the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox is approximately 0.9873 or 98.73%.

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