Three coins are tossed simultaneously. Consider the event $E$ ' three heads or three tails', $\mathrm{F}$ 'at least two heads' and $\mathrm{G}$ ' at most two heads '. Of the pairs $(E,F)$, $(E,G)$ and $(F,G)$, which are independent? which are dependent ?
The sample space of the experiment is given by
Clearly $\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$
$\mathrm{E}=\{\mathrm{HHH}, \mathrm{TTT}\}, \mathrm{F}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$
and $\mathrm{G}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$
Also $\mathrm{E} \cap \mathrm{F}=\{\mathrm{HHH}\}, \mathrm{E} \cap \mathrm{G}=\{\mathrm{TTT}\}, \mathrm{F} \cap \mathrm{G}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$
Therefore $\mathrm{P}(\mathrm{E})=\frac{2}{8}=\frac{1}{4}, \mathrm{P}(\mathrm{F})=\frac{4}{8}=\frac{1}{2}, \mathrm{P}(\mathrm{G})=\frac{7}{8}$
and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{8}, \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{1}{8}, \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{3}{8}$
Also $\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}, \mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{G})=\frac{1}{4} \times \frac{7}{8}=\frac{7}{32}$
and $\mathrm{P}(\mathrm{F}), \mathrm{P}(\mathrm{G})=\frac{1}{2} \times \frac{7}{8}=\frac{7}{16}$
Thus $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F})$
$\mathrm{P}(\mathrm{E} \cap \mathrm{G}) \neq \mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{G})$
and $\mathrm{P}(\mathrm{F} \cap \mathrm{G}) \neq \mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G})$
Hence, the events $(E$ and $F)$ are independent, and the events $(E$ and $G)$ and $(F$ and $G) $ are dependent.
$A , B, C$ try to hit a target simultaneously but independently. Their respective probabilities of hitting targets are $\frac{3}{4},\frac{1}{2},\frac{5}{8}$. The probability that the target is hit by $A$ or $B$ but not by $C$ is
Two cards are drawn at random and without replacement from a pack of $52$ playing cards. Finds the probability that both the cards are black.
If $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are
Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$
Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$
Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{{12}}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then