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The number of food orders at drive-thru centres of four fast-food chains are provided.

Some orders are accurate, while others are inaccurate.

For three events, A, B, and C, to either occur individually or together, the following probability is calculated:

$P\left(AorBorC\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(AandB\right)-P\left(AandC\right)-P\left(BandC\right)$

The subtotals for each chain are tabulated below:

The total number of food orders is equal to 1118.

Let E be the event of selecting a food order from 惭肠顿辞苍补濒诲鈥檚.

Let F be the event of selecting a food order from 奥别苍诲测鈥檚.

Let G be the event of selecting an inaccurate food order.

The number of food orders from 惭肠顿辞苍补濒诲鈥檚 is equal to 362.

The probability of selecting a food order from 惭肠顿辞苍补濒诲鈥檚,

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

.

The number of food orders from 奥别苍诲测鈥檚 is equal to 280.

The probability of selecting a food order from 奥别苍诲测鈥檚,

$P\left(F\right)=\frac{280}{1118}$

The number of inaccurate food orders is equal to 131.

The probability of selecting an inaccurate food order,

$P\left(G\right)=\frac{131}{1118}$

The number of orders from 惭肠顿辞苍补濒诲鈥檚 and 奥别苍诲测鈥檚 is equal to 0. Thus,

$P\left(E\text{and}F\right)=\frac{0}{1118}$

The number of inaccurate orders from 惭肠顿辞苍补濒诲鈥檚 is equal to 33. Thus,

$P\left(E\text{and}G\right)=\frac{33}{1118}$

The number of inaccurate orders from 奥别苍诲测鈥檚 is equal to 31. Thus,

$P\left(F\text{and}G\right)=\frac{31}{1118}$

The probability of getting an order from 惭肠顿辞苍补濒诲鈥檚 or 奥别苍诲测鈥檚 or an inaccurate order is calculated as follows:

$$\begin{gathered} P\left(E\text{or}F\text{or}G\right)=P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(E\text{and}F\right)-P\left(E\text{and}G\right)-P\left(F\text{and}G\right) \\ =\frac{362}{1118}+\frac{280}{1118}+\frac{131}{1118}-\frac{0}{1118}-\frac{33}{1118}-\frac{31}{1118} \\ =\frac{709}{1118} \\ =0.6342 \end{gathered}$$

Therefore, the probability of getting an order from 惭肠顿辞苍补濒诲鈥檚 or 奥别苍诲测鈥檚 or an inaccurate order is equal to 0.634.