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?

Given:
 p = 90% = 0.9, the probability that an adult has had chickenpox by age 50.
Therefore,
q = 1 - p = 0.1, the probability that an adult has not had chickenpox by age 50.

Part (a)
Because there are only two answers: "Yes" or "No" to whether an adult has had chickenpox by age 50, the use of the binomial distribution is justified.

Part (b):
Calculate the probability that exactly 97 out of 100 sampled adults have had chickenpox.
The probability is
P₁ = ₁₀₀C₉₇ (0.9)⁹⁷ (0.1)³ = 0.0059

Answer: 0.006 or 0.6%

Part (c)
Calculate the probability that exactly 3 adults have not had chickenpox.
Theis probability is equal to 
P₂ = 1 - P₁ = 1 - 0.006 = 0.994

Answer: 0.994 or 99.4%

Part (d)
Calculate the probability that at least 1 out of 10 randomly selected adults have had chickenpox.
The probability is
P₃ = ₁₀C₀ (0.9)⁰ (0.1)¹⁰ + ₁₀C₁ (0.9)¹ (0.1)⁹ = 10⁻¹⁰ + 10⁻⁹ = 10⁻⁹ ≈ 0

Answer: 0

Part (e)
Calculate the probability that at most 3 out of 10 randomly selected adults have not had chickenpox.
The probability is
P₄ = 1 - [₁₀C₀ (0.9)⁰(0.1)¹⁰ + ₁₀C₁ (0.9)¹(0.1)⁹ + ₁₀C₂ (0.9)²(0.1)⁸ + ₁₀C₃ (0.9)³(0.1)⁷]
     = 1 - (10⁻¹⁰ + 9 x 10⁻⁹ + 3.645 x 10⁻⁷ + 8.748 x 10⁻⁶)
     = 1

Answer:  1.0 or 100%