Exclusive Or The exclusive or means either one or the other events occurs, but not both.

a. For the formal addition rule, rewrite the formula for P(A or B) assuming that the addition rule uses the exclusive or instead of the inclusive or.

b. Repeat Exercise 11 “Fast Food Drive-Thru Accuracy” using the exclusive or instead of the inclusive or.

The formal addition rule of probability uses the inclusive or rule. It is required to transform the formula using the exclusive or rule.

The formal rule has the following expression:

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

This describes the probability of occurrence of either A or B, or both.

Under this, the probability of occurrence of either A or B, but not both, needs to be expressed.

a.

The original formula of additional rule (inclusive or) is as follows:

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

Here, the probabilities of events A and B include the event where A and B both occur.

To completely remove the probability of occurrence of both A and B from the expression to make it exclusive or, the following expression is devised:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)-2P\left(AandB\right)$

Thus, the new exclusive or additional rule of probability becomes:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)-2P\left(AandB\right)$

b.

The following table is considered for all calculations (as in Exercise 11)

Let E be the event of getting a food order from McDonald’s.

Let F be the event of getting an accurate food order.

The total number of food orders is equal to 1118.

The number of food orders from McDonald’s is calculated as shown below:

$329+33=362$

$P\left(E\right)=\frac{362}{1118}$

The number of accurate food orders is calculated as shown below:

$329+264+249+145=987$

$P\left(F\right)=\frac{987}{1118}$

The number of accurate food orders from McDonald’s is calculated as shown below:

$P\left(EandF\right)=\frac{329}{1118}$

Now, the probability (E or F) using exclusive or becomes:

$$\begin{gathered} P\left(EorF\right)=P\left(E\right)+P\left(F\right)-2P\left(EandF\right) \\ =\frac{362}{1118}+\frac{987}{1118}-2\left(\frac{329}{1118}\right) \\ =\frac{691}{1118} \\ =0.618 \end{gathered}$$

Therefore, the probability of getting a food order from McDonald’s or getting an accurate food order using the exclusive or rule is equal to 0.618.