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The four homeowners are selected at random to check whether they have an earthquake insurance or not. Consider S denote a homemaker who has insurance while F denote a homemaker who do not insurance. The probability of S is 0.25 and for F, it is 0.75.

Consider\(X\)to be the number of homeowners who insured against earthquake. As the homemaker who has earthquake insurance is denoted by S while the homemaker who do not have an insurance is denoted by S. Following are the possible outcomes of random variable\(X\):

\({\rm{Sample}}\,{\rm{space}} = \left\{ \begin{array}{l}\left( {SSSS} \right),\left( {SSSF} \right),\left( {SSFF} \right),\left( {SFFF} \right),\\\left( {SFFS} \right),\left( {SFSF} \right),\left( {SSFS} \right),\left( {SFSS} \right),\\\left( {FSSS} \right),\left( {FFSF} \right),\left( {FSFS} \right),\left( {FSSF} \right),\\\left( {FSSS} \right),\left( {FFSS} \right),\left( {FFFS} \right),\left( {FFFF} \right)\end{array} \right\}\)

When the value of\(X\)is 0 it means no one out of four homeowners have earthquake insurance. There is one outcome i.e. FFFF out of 16 outcomes that fulfil the above condition.

\(\begin{aligned}P\left( {X = 0} \right) &= P\left( {FFFF} \right)\\ &= {\left( {0.75} \right)^4}\\ &= 0.3164\end{aligned}\)

When the value of\(X\)is 1 it means one out of four homeowners have earthquake insurance. There are four outcomes i.e. FFFS, FFSF, FSFF, SFFF out of 16 outcomes that fulfil the above condition.

\(\begin{aligned}P\left( {X = 1} \right) &= P\left( {FFFS} \right) + P\left( {FFSF} \right) + P\left( {FSFF} \right) + P\left( {SFFF} \right)\\ &= \left( \begin{array}{c}{\left( {0.75} \right)^3}\left( {0.25} \right) + {\left( {0.75} \right)^3}\left( {0.25} \right) + {\left( {0.75} \right)^3}\left( {0.25} \right)\\ + {\left( {0.75} \right)^3}\left( {0.25} \right)\end{array} \right)\\ &= 0.1055 \times 4\\ &= 0.4220\end{aligned}\)

When the value of\(X\)is 2 it means two out of four homeowners have earthquake insurance. There are six outcomes i.e. FFSS, SSFF, FSFS, SFSF, FSSF and SFFS out of 16 outcomes that fulfil the above condition.

\(\begin{aligned}P\left( {X = 2} \right) &= \left( \begin{array}{c}P\left( {FFSS} \right) + P\left( {SSFF} \right) + P\left( {FSFS} \right)\\ + P\left( {SFSF} \right) + P\left( {FSSF} \right) + P\left( {SFFS} \right)\end{array} \right)\\ &= \left( \begin{array}{c}{\left( {0.75} \right)^2}{\left( {0.25} \right)^2} + {\left( {0.75} \right)^2}{\left( {0.25} \right)^2} + {\left( {0.75} \right)^2}{\left( {0.25} \right)^2}\\ + {\left( {0.75} \right)^2}{\left( {0.25} \right)^2} + {\left( {0.75} \right)^2}{\left( {0.25} \right)^2} + {\left( {0.75} \right)^2}{\left( {0.25} \right)^2}\end{array} \right)\\ &= 0.03516 \times 6\\ &= 0.2109\end{aligned}\)

When the value of\(X\)is 3 it means three out of four homeowners have earthquake insurance. There are four outcomes i.e. FFSS, SSFF, FSFS, SFSF, FSSF and SFFS out of 16 outcomes that fulfil the above condition.

\(\begin{aligned}P\left( {X = 3} \right) &= \left( \begin{array}{c}P\left( {FSSS} \right) + P\left( {SFSS} \right)\\ + P\left( {SSFS} \right) + P\left( {SSSF} \right)\end{array} \right)\\ &= \left( \begin{array}{l}\left( {0.75} \right){\left( {0.25} \right)^3} + \left( {0.75} \right){\left( {0.25} \right)^3}\\ + \left( {0.75} \right){\left( {0.25} \right)^3} + \left( {0.75} \right){\left( {0.25} \right)^3}\end{array} \right)\\& = 0.01172 \times 4\\ &= 0.0468\end{aligned}\)

When the value of\(X\)is 4 it means four out of four homeowners have earthquake insurance. There is one outcome i.e. SSSS out of 16 outcomes that fulfil the above condition.

\(\begin{aligned}P\left( {X = 4} \right) &= P\left( {SSSS} \right)\\ &= {\left( {0.25} \right)^4}\\ &= 0.0039\end{aligned}\)

The probability distribution of\(X\)is given as,

Following are the steps to construct a probability histogram:

1. Draw two axes in x-y Cartesian plane.
2. The scale of X-axis represent the possible value of\(X\)and the scale of Y-axis represent the pmf of each value.
3. Label the marks and give names to the horizontal and vertical axes.
4. Create bars for each value of pmf w.r.t to the value of\(X\). The height of each bar correspond to the pmf value.

In this, the most likely value of\(X\)is obtained by using the expected value or mean formula as below:

\(E\left( {X = x} \right) = \sum\limits_{x = 0}^4 {xp\left( x \right)} \)

So, the required value is obtained as:

\(\begin{aligned}E\left( {X = x} \right) &= \left( \begin{array}{l}\left( {0 \times 0.3164} \right) + \left( {1 \times 0.4220} \right) + \left( {2 \times 0.2109} \right)\\ + \left( {3 \times 0.0468} \right) + \left( {4 \times 0.0039} \right)\end{array} \right)\\ &= \left( {0 + 0.4220 + 0.4218 + 0.1404 + 0.0156} \right)\\ &= 0.9998\\ \approx 1\end{aligned}\)

Therefore, the most likely value of\(X\)is 1.

In this, the required probability is\(P\left( {X \ge 2} \right)\)as two out of four homeowners have earthquake insurance. It can be computed as,

\(P\left( {X \ge 2} \right) = P\left( {X = 2} \right) + P\left( {X = 3} \right) + P\left( {X = 4} \right)\)

Substitute the value of\(P\left( {X = 2} \right) = 0.2109\),\(P\left( {X = 3} \right) = 0.0468\)and\(P\left( {X = 4} \right) = 0.0039\)in the above formula:

\(\begin{aligned}P\left( {X \ge 2} \right) &= 0.2109 + 0.0468 + 0.0039\\ &= 0.2616\end{aligned}\)

Therefore, the required probability is 0.2616.