91影视

The given data shows the number of accurate and inaccurate food orders at drive-thru centers of four fast food chains.

For three events A, B, and C to occur simultaneously, the following probability is calculated:

$P\left(A\text{and}B\text{and}C\right)=P\left(A\right)脳P\left(B\right)脳P\left(C\right)$

The table below shows the subtotals for each category:

The total number of food orders is equal to 1118.

Let E be the event of selecting an inaccurate food order on the first selection.

Let F be the event of selecting an inaccurate food order on the second selection.

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

The number of inaccurate food orders is equal to 131.

The probability of selecting an inaccurate food order on the first selection is equal to:

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

The remaining total number of orders and the number of inaccurate orders will decrease by 1.

The probability of selecting an inaccurate food order on the second selection is equal to:

$P\left(F\right)=\frac{130}{1117}$

For the third selection, the remaining total number of orders and the number of inaccurate orders will further decrease by 1.

The probability of selecting an inaccurate food order on the third selection is equal to:

$P\left(G\right)=\frac{129}{1116}$

The probability of selecting three inaccurate orders is given by:

$$\begin{gathered} P\left(E\text{and}F\text{and}G\right)=P\left(E\right)脳P\left(F\right)脳P\left(G\right) \\ =\frac{131}{1118}脳\frac{130}{1117}脳\frac{129}{1116} \\ =0.00158 \end{gathered}$$

Therefore, the probability of selecting three inaccurate orders is equal to 0.00158.