The probability that the home team will win an upcoming football game is 0.77 , the probability that it will tie the game is 0.08 and the probability that it will lose the game is ... .
$$P(\text { lossing })=1-(0.77+0.08)=0.15$$
If $e_1, e_2, e_3$ and $e_4$ are the four elementary outcomes in a sample space and $P\left(\mathrm{e}_1\right)=0.1, P\left(\mathrm{e}_2\right)=0.5$ and $P\left(\mathrm{e}_3\right)=0.1$, then the probability of $e_4$ is .... .
$$\begin{array}{lr} \because & P\left(\mathrm{e}_1\right)+P\left(\mathrm{e}_2\right)+P\left(\mathrm{e}_3\right)+P\left(\mathrm{e}_4\right)=1 \\ \Rightarrow & 0.1+0.5+0.1+P\left(\mathrm{e}_4\right)=1 \\ \Rightarrow & 0.7+P\left(\mathrm{e}_4\right)=1 \\ \therefore & P\left(\mathrm{e}_4\right)=0.3 \end{array}$$
$$ \text { If } S=\{1,2,3,4,5,6\} \text { and } E=\{1,3,5\} \text {, then } \bar{E} \text { is } \ldots . $$
$$\begin{aligned} \text{Here,}\quad & S=\{1,2,3,4,5,6\} \\ \text{and}\quad & E=\{1,3,5\} \\ \therefore\quad& \bar{E}=S-E=\{2,4,6\} \end{aligned}$$
If $A$ and $B$ are two events associated with a random experiment such that $P(A)=0.3, P(B)=0.2$ and $P(A \cap B)=0.1$, then the value of $P(A \cap \bar{B})$ is ... .
$$P(A \cap \bar{B})=P(A)-P(A \cap B)=0.3-0.1=0.2$$
The probability of happening of an event $A$ is 0.5 and that of $B$ is 0.3 . If $A$ and $B$ are mutually exclusive events, then the probability of neither $A$ nor $B$ is ....
$$\begin{aligned} P(\bar{A} \cap \bar{B})=P(\overline{A \cup B}) & =1-P(A \cup B) \\ & =1-[P(A)+P(B)] \quad \text { [since, } A \text { and } B \text { are mutually exclusive] } \\ & =1-(0.5+0.3)=1-0.8=0.2 \end{aligned}$$