Answer with Step-by-step explanation:
S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),..,(2,6),(3,1),(3,2),...,(3,6),(4,1),..,(4,6),(5,1),(5,2),(5,3),...,(5,6),(6,1),(6,2),..,(6,6)}
Total number of cases=36
M=Maximum of the two tosses
M=1,
(1,1)
Therefore, [tex]P(1)=\frac{1}{36}[/tex]
Using the formula, P(E)=[tex]\frac{favorable\;cases}{Total\;number\;of\;cases}[/tex]
M=2
(1,2),(2,1),(2,2)
Favorable cases=3
[tex]P(2)=\frac{3}{36}=\frac{1}{12}[/tex]
M=3
(1,3),(3,1),(3,2),(2,3),(3,3)
[tex]P(3)=\frac{5}{36}[/tex]
M=4
(1,4),(4,1),(2,4),(3,4),(4,2)(4,3),(4,4)
[tex]P(4)=\frac{7}{36}[/tex]
M=5
(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)
[tex]P(5)=\frac{9}{36}=\frac{1}{4}[/tex]
M=6
(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
[tex]P(6)=\frac{11}{36}[/tex]
