91Ó°ÊÓ

The number of food orders from four fast-food chains is provided.

The events are defined as follows:

A: getting an order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s and B: getting an order from Burger King.

The addition rule has the following expression:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)-P\left(AandB\right)$

This represents the probability of occurrence of either A or B, or both.

A and $\stackrel{¯}{A}$are said to be complementary events when the former signifies the occurrence of an event while the latter is the non-occurrence of the same event.

$P\left(A\right)+P\left(\stackrel{¯}{A}\right)=1$

The following table is considered for all calculations (as in Exercise 9 to 20)

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

Let B be the event of getting a food order from Burger King.

The probability of getting a food order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s is equal to:

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

The probability of getting a food order from Burger King is equal to:

$P\left(B\right)=\frac{318}{1118}$

The number of food orders from Burger King as well as McDonald's is equal to 0.

The probability of getting a food order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s and Burger King is given as:

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

Using the complement rule, the required probability is expressed as follows:

$$\begin{gathered} P\left(\stackrel{¯}{AorB}\right)=1-P\left(AorB\right) \\ =1-\left[P\left(A\right)+P\left(B\right)-P\left(AorB\right)\right] \\ =1-\left[\frac{362}{1118}+\frac{318}{1118}-\frac{0}{1118}\right] \\ =0.392 \end{gathered}$$

Therefore,$P\left(\stackrel{¯}{AorB}\right)=0.392$

The probability of not getting a food order from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s is equal to:

$$\begin{gathered} P\left(\stackrel{¯}{A}\right)=1-\frac{362}{1118} \\ =\frac{756}{1118} \end{gathered}$$

The probability of not getting a food order from Burger King is equal to:

$$\begin{gathered} P\left(\stackrel{¯}{B}\right)=1-\frac{318}{1118} \\ =\frac{800}{1118} \end{gathered}$$

The number of food orders that are neither from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s nor Burger King is equal to the sum of orders from the remaining two chains.

$280+158=438$

The probability of getting a food order neither from ²Ñ³¦¶Ù´Ç²Ô²¹±ô»å’s nor Burger King is equal to:

$$\begin{gathered} P\left(\stackrel{¯}{A}or\stackrel{¯}{B}\right)=P\left(\stackrel{¯}{A}\right)+P\left(\stackrel{¯}{B}\right)-P\left(\stackrel{¯}{A}and\stackrel{¯}{B}\right) \\ =\frac{756}{1118}+\frac{800}{1118}-\frac{438}{1118} \\ =1 \end{gathered}$$

Therefore, $P\left(\stackrel{¯}{A}or\stackrel{¯}{B}\right)≈1$.

The numerical values of the two probabilities are unequal.

Thus, it cannot be concluded for the general setup that $P\left(\stackrel{¯}{AorB}\right)=P\left(\stackrel{¯}{A}or\stackrel{¯}{B}\right)$.