$$ \text { If } S=\{1,2,3,4,5,6\} \text { and } E=\{1,3,5\} \text {, then } \bar{E} \text { is } \ldots . $$
$$\begin{aligned} \text{Here,}\quad & S=\{1,2,3,4,5,6\} \\ \text{and}\quad & E=\{1,3,5\} \\ \therefore\quad& \bar{E}=S-E=\{2,4,6\} \end{aligned}$$
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 ....
$$\begin{aligned} P(\bar{A} \cap \bar{B})=P(\overline{A \cup B}) & =1-P(A \cup B) \\ & =1-[P(A)+P(B)] \quad \text { [since, } A \text { and } B \text { are mutually exclusive] } \\ & =1-(0.5+0.3)=1-0.8=0.2 \end{aligned}$$
Match the following.
| Column I | Column II | ||
|---|---|---|---|
| (i) | 0.95 | (a) | An incorrect assignment |
| (ii) | 0.02 | (b) | No chance of happening |
| (iii) | $-$0.3 | (c) | As much chance of happening as not |
| (iv) | 0.5 | (d) | Very likely to happen |
| (v) | 0 | (e) | Very little chance of happening |
(i) 0.95 is very likely to happen, so it is close to 1.
(ii) 0.02 very little chance of happening because probability is very low.
(iii) $-$0.3 an incorrect assignment because probability of any events lie between 0 and 1.
(iv) 0.5 , as much chance of happening as not because sum of chances of happening and not happening is zero.
(v) 0, no chance of happening.
Match the following.
| Column I | Column II | ||
|---|---|---|---|
| (i) | If $E_1$ and $E_2$ are the two mutually exclusive events | (a) | $E_1 \cap E_2=E_1$ |
| (ii) | If $E_1$ and $E_2$ are the mutually exclusive and exhaustive events | (b) | $\left(E_1-E_2\right) \cup\left(E_1 \cap E_2\right)=E_1$ |
| (iii) | If $E_1$ and $E_2$ have common outcomes, then | (c) | $E_1 \cap E_2=\phi, E_1 \cup E_2=S$ |
| (iv) | If $E_1$ and $E_2$ are two events such that $E_1 \subset E_2$ | (d) | $E_1 \cap E_2=\phi$ |
(i) If $E_1$ and $E_2$ are two mutually exclusive event, then $E_1 \cap E_2=\phi$.
(ii) If $E_1$ and $E_2$ are mutually exclusive and exhaustive events, then $E_1 \cap E_2=\phi$ and $E_1 \cup E_2=S$.
(iii) If $E_1$ and $E_2$ have common outcomes, then $\left(E_1-E_2\right) \cup\left(E_1 \cap E_2\right)=E_1$
(iv) If $E_1$ and $E_2$ are two events such that $E_1 \subset E_2 \Rightarrow E_1 \cap E_2=E_1$
