Chapter 4, Problem 4.3.20RE

To determine

The probability that all four labels are applied correctly, exactly three labels are correct, exactly two labels are correct and at least one label is correct.

Answer to Problem 4.3.20RE

The probability that all four labels are applied correctly has $\frac{1}{256}$ probability.

The probability that exactly three labels were applied correctly has $\frac{3}{64}$ probability.

The probability that exactly two labels were applied correctly has $\frac{27}{128}$ probability.

The probability that at least one label was applied correctly has $\frac{175}{256}$ probability.

Explanation of Solution

Given info:

There are four packages without shipping labels, a person applies labels for all the four packages.

Calculation:

The possible ways to apply the labels on four packages are given below:

$S=\left\{\begin{array}{l}CCCC,CCCI,CCIC,CCII,CICC,CICI,CIIC,CIII,\\ ICCC,ICCI,ICIC,ICII,IICI,IICC,IIIC,IIII\end{array}\right\}$

Where, C represents the correct placement of labels and I represents the incorrect placement of labels.

The probability that a label is correct has $\frac{1}{4}$ probability.

The probability that a label is incorrect has $\frac{3}{4}$ probability.

The probability that all four labels were applied correctly is given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{All four labels were}\\ \text{applied correctly}\end{array}\right)=\left(\begin{array}{l}\text{Number of ways to apply }\\ \text{all four labels correctly}\end{array}\right)\times P\left(\text{Label is applied correctly}\right)\\ =1{\left(\frac{1}{4}\right)}^4\\ =\frac{1}{256}\end{array}$

Thus, the probability that all four labels were applied correctly is $\frac{1}{256}$

The probability that exactly three labels were applied correctly is given below:

Let A be the event of applying exactly three correct labels,

$A=\left\{CCCI,CCIC,CICC,ICCC\right\}$

There are 4 ways to apply exactly three correct labels

The probability that exactly three labels were applied correctly is

$\begin{array}{c}P\left(\begin{array}{l}\text{Exactly three }\\ \text{correct labels}\end{array}\right)=\left(\begin{array}{l}\text{Number of ways to apply }\\ \text{exactly three correct labels}\end{array}\right)\times P\left(\begin{array}{l}\text{Applying three }\\ \text{correct label}\end{array}\right)\times P\left(\begin{array}{l}\text{Applying a }\\ \text{incorrect label}\end{array}\right)\\ =4\left(\frac{1}{4}\right)\times \left(\frac{1}{4}\right)\times \left(\frac{1}{4}\right)\times \left(\frac{3}{4}\right)\\ =\frac{3}{64}\end{array}$

Thus, the probability that exactly three labels were applied correctly is $\frac{3}{64}$

The probability that exactly two labels were applied correctly is given below:

Let B be the event of applying exactly two correct labels,

$B=\left\{CCII,CICI,CIIC,ICCI,IICC,ICIC\right\}$

There are 6 ways to apply exactly two correct labels

The probability that exactly two labels were applied correctly is

$\begin{array}{c}P\left(\begin{array}{l}\text{Exactly two }\\ \text{correct labels}\end{array}\right)=\left(\begin{array}{l}\text{Number of ways to apply }\\ \text{exactly two correct labels}\end{array}\right)\times P\left(\begin{array}{l}\text{Applying two }\\ \text{correct label}\end{array}\right)\times P\left(\begin{array}{l}\text{Applying two }\\ \text{incorrect label}\end{array}\right)\\ =6\left(\frac{1}{4}\right)\times \left(\frac{1}{4}\right)\times \left(\frac{3}{4}\right)\times \left(\frac{3}{4}\right)\\ =\frac{27}{128}\end{array}$

Thus, the probability that exactly two labels were applied correctly is $\frac{27}{128}$

The probability that at leastone label was applied correctly is given below:

$\begin{array}{c}P\left(\text{At least one correct label}\right)=1-P\left(\text{All the four labels are incorrect}\right)\\ =1-{\left(\frac{3}{4}\right)}^4\\ =1-\frac{81}{256}\\ =\frac{175}{256}\end{array}$

Thus, the probability that at least one label was applied correctly is $\frac{175}{256}$