To solve this problem, let's carefully analyze the given probability distribution and the conditions required for parts (a) and (b). The probability distribution table is as follows:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Value \( x \) of \( X \) & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
\( P(X = x) \) & 0.20 & 0.23 & 0.28 & 0.16 & 0.10 & 0.03 \\
\hline
\end{tabular}
\][/tex]
### Part (a): Determine the probability that the number of children is greater than 2.
We are interested in finding [tex]\( P(X > 2) \)[/tex]. This involves summing the probabilities of having more than 2 children, i.e., the probabilities for [tex]\( X = 3 \)[/tex], [tex]\( X = 4 \)[/tex], and [tex]\( X = 5 \)[/tex].
[tex]\[
P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)
\][/tex]
Given the values from the distribution,
[tex]\[
P(X = 3) = 0.16, \quad P(X = 4) = 0.10, \quad P(X = 5) = 0.03
\][/tex]
Sum these probabilities:
[tex]\[
P(X > 2) = 0.16 + 0.10 + 0.03 = 0.29
\][/tex]
Thus, the probability that the number of children is greater than 2 is [tex]\(\boxed{0.29}\)[/tex].
### Part (b): Determine the probability that the number of children is less than 3.
We are interested in finding [tex]\( P(X < 3) \)[/tex]. This involves summing the probabilities of having less than 3 children, i.e., the probabilities for [tex]\( X = 0 \)[/tex], [tex]\( X = 1 \)[/tex], and [tex]\( X = 2 \)[/tex].
[tex]\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\][/tex]
Given the values from the distribution,
[tex]\[
P(X = 0) = 0.20, \quad P(X = 1) = 0.23, \quad P(X = 2) = 0.28
\][/tex]
Sum these probabilities:
[tex]\[
P(X < 3) = 0.20 + 0.23 + 0.28 = 0.71
\][/tex]
Thus, the probability that the number of children is less than 3 is [tex]\(\boxed{0.71}\)[/tex].
