The probability of intersection of two events $A$ and $B$ is always less than or equal to those favourable to the event $A$.
The probability of an occurrence of event $A$ is 0.7 and that of the occurrence of event $B$ is 0.3 and the probability of occurence of both is 0.4 .
The sum of probabilities of two students getting distinction in their final examinations is 1.2 .
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 .... .
$$\begin{array}{lr} \because & P\left(\mathrm{e}_1\right)+P\left(\mathrm{e}_2\right)+P\left(\mathrm{e}_3\right)+P\left(\mathrm{e}_4\right)=1 \\ \Rightarrow & 0.1+0.5+0.1+P\left(\mathrm{e}_4\right)=1 \\ \Rightarrow & 0.7+P\left(\mathrm{e}_4\right)=1 \\ \therefore & P\left(\mathrm{e}_4\right)=0.3 \end{array}$$