Alright, let's solve part (a) of the question step-by-step.
We are given the probability distribution for the number of customers in line at a supermarket express checkout counter. The variable [tex]\( x \)[/tex] represents the number of customers, and [tex]\( P(x) \)[/tex] represents the probability of [tex]\( x \)[/tex] customers being in line.
The probability distribution is provided as follows:
[tex]\[
\begin{array}{c|cccccc}
x & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
P(x) & 0.05 & 0.25 & 0.30 & 0.25 & 0.10 & 0.05 \\
\end{array}
\][/tex]
To find [tex]\( P(1) \)[/tex], we need to look at the corresponding probability value when [tex]\( x = 1 \)[/tex].
From the table, we can see that:
[tex]\[ P(1) = 0.25 \][/tex]
Therefore, the probability that there is exactly 1 customer in line at the express checkout counter is:
[tex]\[ P(1) = 0.25 \][/tex]
