**Toolbox:**

- In a pair of die thrown together total number of events 6x6=36
- E=getting a total of 4
- E=\({(1,3),(3,1),(2,2)}\)
- F=Total of 9or more
- \({(3,6),(6,3),(4,5),(5,4),(5,5)(6,4),(4,6),(5,6)(6,5),(6,6)}\)
- G=Total disable by 5
- Total of 5or10
- \({(1,4),(4,1),(2,3),(3,2),(5,5),(6,4),(4,6)}\)
- E\(\cap\)F=0
- F\(\cap\)G=\({(5,5),(6,4),(4,6)}\)
- E\(\cap\)G=0
- EventA and B are indenpent if P(A)P(B)=P(A\(\cap\)B)

P(E)=\(\Large\frac{3}{36}\)=\(\Large\frac{1}{12}\)

P(F)=\(\Large\frac{10}{36}\)

P(G)=\(\Large\frac{7}{36}\)

P(E\(\cap\)F)=0e

P(E\(\cap\)G)=0

P(F\(\cap\)G)=\(\Large\frac{3}{36}\)=\(\Large\frac{1}{12}\)

P(E)P(F)=\(\Large\frac{10}{36}\times\)\(\Large\frac{1}{12}\)

=\(\Large\frac{5}{216}\)\(\neq \) P(E\(\cap\)F)

P(E)P(G)=\(\Large\frac{1}{12}\)x\(\Large\frac{7}{36}\)\(\neq\)p(E\(\cap\)G)

EandG arenot independent

P(F)P(G)=\(\Large\frac{10}{36}\)x\(\Large\frac{7}{36}\)\(\neq\)P(F\(\cap\)G)

F and G are not independent