Answer:
(a) The probability mass function of <em>X</em> is:
![P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7B4%5Cchoose%20x%7D%5C%20%280.33%29%5E%7Bx%7D%5C%20%281-0.33%29%5E%7B4-x%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
(b) The most likely value for <em>X</em> is 1.32.
(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.
Step-by-step explanation:
The random variable <em>X</em> is defined as the number among the four homeowners who have earthquake insurance.
The probability that a homeowner has earthquake insurance is, <em>p</em> = 0.33.
The random sample of homeowners selected is, <em>n</em> = 4.
The event of a homeowner having an earthquake insurance is independent of the other three homeowners.
(a)
All the statements above clearly indicate that the random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 4 and <em>p</em> = 0.33.
The probability mass function of <em>X</em> is:
![P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7B4%5Cchoose%20x%7D%5C%20%280.33%29%5E%7Bx%7D%5C%20%281-0.33%29%5E%7B4-x%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
(b)
The most likely value of a random variable is the expected value.
The expected value of a Binomial random variable is:
![E(X)=np](https://tex.z-dn.net/?f=E%28X%29%3Dnp)
Compute the expected value of <em>X</em> as follows:
![E(X)=np](https://tex.z-dn.net/?f=E%28X%29%3Dnp)
![=4\times 0.33\\=1.32](https://tex.z-dn.net/?f=%3D4%5Ctimes%200.33%5C%5C%3D1.32)
Thus, the most likely value for <em>X</em> is 1.32.
(c)
Compute the probability that at least two of the four selected have earthquake insurance as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
![=1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015](https://tex.z-dn.net/?f=%3D1-%7B4%5Cchoose%200%7D%5C%20%280.33%29%5E%7B0%7D%5C%20%281-0.33%29%5E%7B4-0%7D-%7B4%5Cchoose%201%7D%5C%20%280.33%29%5E%7B1%7D%5C%20%281-0.33%29%5E%7B4-1%7D%5C%5C%5C%5C%3D1-0.20151121-0.39700716%5C%5C%5C%5C%3D0.40148163%5C%5C%5C%5C%5Capprox%200.4015)
Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.