Question

Two fair dice are rolled and the sum r of the numbers turned up is considered

• A. $Pr(r>6)=\frac{1}{6}$
• B. $Pr\left(r/3isaninteger\right)=\frac{5}{6}$
• C. $Pr(r=8/(r/4\left)isaninteger\right)=\frac{5}{9}$
• D. $Pr(r=6/(r/5\left)isaninteger\right)=\frac{1}{18}$

Answer 1

The correct option is C $Pr(r=8/(r/4\left)isaninteger\right)=\frac{5}{9}$

P(2)	P(3)	P(4)	P(5)	P(6)	P(7)
$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$
P(12)	P(11)	P(10)	P(9)	P(8)

P(r > 6) = P(7, 8, 9, 10, 11, 12)

$\therefore$ P(r > 6) = P(7) + P(8) + P(9) + P(10) + P(11) + P(12)

$P(r>6)=\frac{1}{36}[6+5+4+3+2+1]$

$=\frac{21}{36}=\frac{7}{12}$

P(r/3 is an integer)
= P(3,6,9,12)

$=\frac{1}{36}[2+5+4+1]=\frac{1}{3}$

P(r/4 is an integer)
= P(4, 8, 12)

$=\frac{1}{36}[3+5+1]=\frac{1}{4}$

$\therefore P\left(8/\right(r/4\left)isaninteger\right)$

$=\frac{P(8\cap (r/4\left)isinteger\right)}{P\left(r/4isinteger\right)}$

$=\frac{5/36}{9/36}=\frac{5}{9}$

P(r/5 is an integer) = P(5, 10)

$=\frac{1}{36}[4+3]=\frac{7}{36}$

P(r/6 / r/5 is an integer)

$=\frac{P(r/6isaninteger\cap r/5isaninteger)}{P\left(r/5isaninteger\right)}$

$=\frac{0}{7/36}=0$