91Ó°ÊÓ

Frequencies of food orders from four different chains are categorized under two subheadings, accurate and inaccurate orders.

Let A and B be two events.

Theprobability of occurrence of either event A, B,or bothhas the following expression:

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

Here, denotes the probability of occurrence of A and B together. This probability needs to be subtracted under the addition rule.

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

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

Let F be the event of getting a food order from °Â±ð²Ô»å²â’s.

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

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

$33+54+31+13=131$

The number of food orders from °Â±ð²Ô»å²â’s is calculated as shown below:

$249+31=280$

The number of inaccurate food orders from °Â±ð²Ô»å²â’s is equal to 31.

So, the required probabilities are the following:

$$\begin{gathered} P\left(E\right)=\frac{131}{1118} \\ P\left(F\right)=\frac{280}{1118} \\ P\left(EandF\right)=\frac{31}{1118} \end{gathered}$$

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

$$\begin{gathered} P\left(EorF\right)=P\left(E\right)+P\left(F\right)-P\left(EandF\right) \\ =\frac{131}{1118}+\frac{280}{1118}-\frac{31}{1118} \\ =\frac{380}{1118} \\ =0.340 \end{gathered}$$

Therefore, the probability of getting a wrong order or an order from °Â±ð²Ô»å²â’s is0.340.

Two events are said to be disjoint if theycannot occur simultaneously.

Since the event of getting a food order from °Â±ð²Ô»å²â’s can occur simultaneously with the event of getting a wrong order, the given two events are not disjoint.

Mathematically, disjoint events E and F have the following probability 0.

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

As, $P\left(EandF\right)=\frac{31}{1118}\left(\text{Non - zero}\right)$, the events are disjoint.