Question

Two fair dice are thrown, find the probability that sum of the points on their uppermost faces is greater than $10$ or an odd number:

• A. $\frac{16}{36}$
• B. $\frac{17}{36}$
• C. $\frac{18}{36}$
• D. $\frac{19}{36}$

Answer 1

Answer: (D)

$\frac{19}{36}$

two fair dice thrown
so total cases$=6\times 6=36$
and for greater than $10$ or and odd no. that mean that sum should be equal to $=3,5,7,9,11,12$
for getting $3=(1,2);(2,1)$
for getting $5=(1,4);(4,1);(2,3);(3,2)$
for getting $7=(1,6);(2,5);(3,4);(4,3);(5,2);(6,1)$
for getting $9=(3,6);(4,5);(5,4);(6,3)$
for getting $11=(5,6);(6,5)$
for getting $12=(6,6)$
so total cases$=19$
so probability$=19/36$