91影视

Let B be the event of selecting a person with type B blood.

If A represents the occurrence of an event and $\stackrel{炉}{A}$represents the occurrence of 鈥渘ot A,鈥� then A and$\stackrel{炉}{A}$ arecomplementary events.

The outcomes that do not belong to event A belong to the complementary set and vice-versa. Consequently, only one of the two events can occur at one point in time.

Moreover, one of the two events is sure to occur.

Mathematically,

$$\begin{gathered} P\left(Aor\stackrel{炉}{A}\right)=P\left(A\right)+P\left(\stackrel{炉}{A}\right) \\ =1 \end{gathered}$$

Here, B is the event of selecting a person with type B blood. Then, is the event of selecting a person whose blood type is not B.

Thus, the probability that either of the two events will occur is given as:

$$\begin{gathered} P\left(Bor\stackrel{炉}{B}\right)=P\left(B\right)+P\left(\stackrel{炉}{B}\right)-P\left(Band\stackrel{炉}{B}\right)...\left(\text{Addition}\text{rule}\right) \\ =P\left(B\right)+P\left(\stackrel{炉}{B}\right)-0 \\ =P\left(B\right)+P\left(\stackrel{炉}{B}\right) \\ =1 \end{gathered}$$

This implies that events B and$\stackrel{炉}{B}$ cannot occur together, and either of the two events will certainly occur.

Specifically, the probability that a person either has blood type B or does not have blood type B is sure to occur for any randomly selected adult.