91Ó°ÊÓ

The given data depicts the number of food orders from four different outlets under two categories, accurate order and inaccurate order, in a tabulated form.

For two events, A and B, the probability is that either event A, B or both occur. It has the following expression:

$P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{and}B\right)$

Two events are said to be disjoint if theyboth cannot occur together.

That is,

$P\left(AandB\right)=0$

The following table shows all the subtotals and the grand total:

Let E be the event of getting a food order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s.

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

The total number of food orders is equal to 1,118.

The number of food orders from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s is calculated as shown below:

$329+33=362$

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

$329+264+249+145=987$

The number of accurate food orders from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s is 329.

So, the required probabilities are obtained as shown below:

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

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

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

Now, the probability (E or F) is as follows:

$$\begin{gathered} P\left(EorF\right)=P\left(E\right)+P\left(F\right)-P\left(EandF\right) \\ =\frac{362}{1118}+\frac{987}{1118}-\frac{329}{1118} \\ =\frac{1020}{1118} \\ =0.9123 \end{gathered}$$

Therefore, the probability of getting a food order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s or getting an accurate food order is equal to0.912.

Since the event of getting a food order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s can occur simultaneously with the event of getting an accurate order, the given two events are not disjoint.

Mathematically,

$$\begin{gathered} P\left(EandF\right)=\frac{329}{1118} \\ ≠0 \end{gathered}$$

Thus, the events are not disjoint.