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The probability that any child in a certain family will have blue eyes is 1/4, and this feature is inherited independently by different children in the family.

There are five children in the family, and it is known that at least one of these children has blue eyes.

The given problem can be modeled with a Binomial distribution.

Let X = the number of children having blue eyes

X follows a binomial distribution with parameters n = 5 and p = 1/4.

The probability of exactly x number of successes out of total n trials of a binomial experiment is termed binomial probability and is given by:

\({\bf{P}}\left( {{\bf{X = x}}} \right){\bf{ = }}{}^{\bf{n}}{{\bf{C}}_{\bf{x}}}{\left( {\bf{p}} \right)^{\bf{x}}}{\left( {{\bf{1}} - {\bf{p}}} \right)^{{\bf{n}} - {\bf{x}}}}{\bf{;x = 0,1,}}...{\bf{,n}}\)

If there are five children in the family and it is known that at least one of these children has blue eyes\(\left( {X \ge 1} \right)\), the probability that at least three of the children have blue eyes\(\left( {X \ge 3} \right)\)is obtained as:

\(\begin{aligned}{}P\left[ {\left( {X \ge 3} \right)|\left( {X \ge 1} \right)} \right] = \frac{{P\left[ {\left( {X \ge 3} \right) \cap \left( {X \ge 1} \right)} \right]}}{{P\left( {X \ge 1} \right)}}\\ = \frac{{P\left( {X \ge 3} \right)}}{{P\left( {X \ge 1} \right)}}\;\;\; - \left( 1 \right)\end{aligned}\)

Using the binomial probability formula, obtaining the probabilities as:

\(\begin{aligned}{}P\left( {X \ge 3} \right) &= P\left( {X = 3} \right) + P\left( {X = 4} \right) + P\left( {X = 5} \right)\\ &= {}^5{C_3}{\left( {\frac{1}{4}} \right)^3}{\left( {1 - \frac{1}{4}} \right)^{5 - 3}} + {}^5{C_4}{\left( {\frac{1}{4}} \right)^4}{\left( {1 - \frac{1}{4}} \right)^{5 - 4}} + {}^5{C_5}{\left( {\frac{1}{4}} \right)^5}{\left( {1 - \frac{1}{4}} \right)^{5 - 5}}\\ &= 0.08789 + 0.01464 + 0.00097\\ &= 0.1035\end{aligned}\)

And,

\(\begin{aligned}{}P\left( {X \ge 1} \right)& = 1 - P\left( {X = 0} \right)\\& = 1 - {}^5{C_0}{\left( {\frac{1}{4}} \right)^0}{\left( {1 - \frac{1}{4}} \right)^{5 - 0}}\\ &= 1 - 0.2373\\ &= 0.7627\end{aligned}\)

Substituting the values in equation (1),

\(\begin{aligned}{}P\left[ {\left( {X \ge 3} \right)|\left( {X \ge 1} \right)} \right] = \frac{{P\left( {X \ge 3} \right)}}{{P\left( {X \ge 1} \right)}}\\ = \frac{{0.1035}}{{0.7627}}\\ = 0.1357\end{aligned}\)

Therefore, the required probability is approximately 0.1357.