Question

Through observation, it has been determined that the probability for a given number of people waiting in line at the “5 items or less” checkout register of a supermarket is as follows:

Find the probability of:

(a) At most 2 people in line

(b) At least 2 people in line

(c) At least 1 person in line

Short answer

The required probabilities are
$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)0.45 \\ \mathrm{Part}\left(\mathrm{b}\right)0.75 \\ \mathrm{Part}\left(\mathrm{c}\right)0.90 \end{gathered}$$

Step-by-step solution

Step 1. Given Information

The given table is

Part(a) Step 1. Calculation

Probability of at most 2 people in line is as follows,

$$\begin{gathered} P\left(E\right)=P\left(0\right)+P\left(1\right)+P\left(2\right) \\ =0.1+0.15+0.20 \\ =0.45 \end{gathered}$$

Part(b) Step 1. Calculation

Probability of at least 2 people in line is as follows,

$$\begin{gathered} P\left(E_1\right)=P\left(2\right)+P\left(3\right)+P(4ormore) \\ =0.20+0.24+0.31 \\ =0.75 \end{gathered}$$

Part(c) Step 1. Calculation

Probability of at least 1 people in line is as follows,

$$\begin{gathered} P\left(E_2\right)=P\left(1\right)+P\left(2\right)+P\left(3\right)+P(4ormore) \\ =0.15+0.20+0.24+0.31 \\ =0.90 \end{gathered}$$