Two dice are thrown together and the total score is noted. The events $E$, $F$ and $G$ are 'a total of 4’, 'a total of 9 or more' and 'a total divisible by 5', respectively. Calculate $P(E),P(F)$ and $P(G)$ and decide which pairs of events, if any are independent.

Two dice are thrown together i.e., sample space $(S) = 36 \Rightarrow n(S) = 36$

$E = A$ total of $4 = \{ (2,2),(3,1),(1,3)\}$
$\Rightarrow$ $n(E) = 3$

$F = A$ total of 9 or more

$= \{ (3,6),(6,3),(4,5),(4,6),(5,4),(6,4),(5,5),(5,6),(6,5),(6,6)\}$
$\Rightarrow$ $n(F) = 10$

$G = a$ total divisible by $5 = \{ (1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)\}$

$\Rightarrow$ $n(G) = 7$
Here, $(E \cap F) = \phi$

and $(E \cap G) = \phi$
Also, $(F \cap G) = \{ (4,6),(6,4),(5,5)\}$

$\Rightarrow$ $n(F \cap G) = 3$

and $(E \cap F \cap G) = \phi$
$\therefore$ $P(E) = \frac{{n(E)}}{{n(S)}} = \frac{3}{{36}} = \frac{1}{{12}}$

$P(F) = \frac{{n(F)}}{{n(S)}} = \frac{{10}}{{36}} = \frac{5}{{18}}$

$P(G) = \frac{{n(G)}}{{n(S)}} = \frac{7}{{36}}$
$P(F \cap G) = \frac{3}{{36}} = \frac{1}{{12}}$

and $P(F) \cdot P(G) = \frac{5}{{18}} \cdot \frac{7}{{36}} = \frac{{35}}{{648}}$

Here, we see that $P(F \cap G) \ne P(F) \cdot P(G)$

[since, only $F$ and $G$ have common events, so only $F$ and $G$ are used here]

Hence, there is no pair which is independent.