Chapter 13.3, Problem 68AYU

To determine

To find: The probability of,

a. At most 2 people in line.

Answer to Problem 68AYU

a. $0.45$

Explanation of Solution

Given:

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:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

a. $P\left(Atmost2people\right)=P\left(0\right)+P\left(1\right)+P\left(2\right)$

$=0.10+0.15+0.2=\text{ 0}.45$

To determine

To find: The probability of,

b. At least 2 people in line.

Answer to Problem 68AYU

b. $0.75$

Explanation of Solution

Given:

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:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

b. $P\left(atleast2people\right)=P\left(2\right)+P\left(3\right)+P\left(4ormore\right)$

$=0.20+0.24+0.31$

$=\text{ 0}.75$

To determine

To find: The probability of,

c. At least 1 person in line.

Answer to Problem 68AYU

c. $0.90$

Explanation of Solution

Given:

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:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

c. $P\left(atleast1person\right)=P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4 or more\right)$

$=0.15+0.20+0.24+0.31$

$=\text{ 0}.90$