91Ó°ÊÓ

The given data shows the number of food orders at drive-thru centers of four fastfood chains.

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

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

The table below shows the subtotals for each category:

The total number of food orders is equal to 1118.

Define the events.

Let E be the event of selecting a food order from °Â±ð²Ô»å²â’s on the first selection.

Let Fbe the event of selecting a food order from °Â±ð²Ô»å²â’s on the second selection.

Let G be the event of selecting a food order from °Â±ð²Ô»å²â’s on the third selection.

The number of food orders from °Â±ð²Ô»å²â’s is equal to 318.

The probability of selecting a food order from °Â±ð²Ô»å²â’s on the first selection is equal to:

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

As the remaining total number of orders and the remaining number of orders from °Â±ð²Ô»å²â’s will be decreased by 1, the probability of selecting a food order from °Â±ð²Ô»å²â’s on the second selection is equal to:

$P\left(F\right)=\frac{279}{1117}$

Further, the remaining total number of orders and the remaining number of orders from °Â±ð²Ô»å²â’s will again decrease by 1. Thus, the probability of selecting a food order from °Â±ð²Ô»å²â’s on the third selection is equal to:

$P\left(G\right)=\frac{278}{1116}$

The probability of selecting three orders from °Â±ð²Ô»å²â’s is given by:

$$\begin{gathered} P\left(Ea\text{nd}F\text{and}G\right)=P\left(E\right)×P\left(F\right)×P\left(G\right) \\ =\frac{280}{1118}×\frac{279}{1117}×\frac{276}{1116} \\ =0.0156 \end{gathered}$$

Therefore, the probability of selecting three orders from °Â±ð²Ô»å²â’s is equal to 0.0156.