Question

Two fair six-sided dice are tossed independently. Let M 5 the maximum of the two tosses (so M(1,5) 5 5, M(3,3) 5 3, etc.). a. What is the pmf of M? [Hint: First determine p(1),

Answer 1

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, $P(1)=\frac{1}{36}$

Using the formula, P(E)=$\frac{favorable\;cases}{Total\;number\;of\;cases}$

M=2

(1,2),(2,1),(2,2)

Favorable cases=3

$P(2)=\frac{3}{36}=\frac{1}{12}$

M=3

(1,3),(3,1),(3,2),(2,3),(3,3)

$P(3)=\frac{5}{36}$

M=4

(1,4),(4,1),(2,4),(3,4),(4,2)(4,3),(4,4)

$P(4)=\frac{7}{36}$

M=5

(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)

$P(5)=\frac{9}{36}=\frac{1}{4}$

M=6

(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

$P(6)=\frac{11}{36}$