#### Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases = 36
case of getting sum 2 = ( 1 , 1 ) ( 1 , 1 )
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 2 =  $\frac{2}{36}= \frac{1}{18}$
case of getting sum 3 = (1, 2), (1, 2), (2, 1), (2, 1)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 3=  $\frac{4}{36}= \frac{1}{9}$
case of getting sum 4 = (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 4=  $\frac{6}{36}= \frac{1}{6}$
case of getting sum 5 = (2, 3), (2, 3), (4, 1),(4,1) (3, 2), (3, 2)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 5 = $\frac{6}{36}= \frac{1}{6}$
case of getting sum 6 = (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 6=  $\frac{6}{36}= \frac{1}{6}$
case of getting sum 7 = (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 7=  $\frac{6}{36}= \frac{1}{6}$
case of getting sum 8 = (5, 3), (5, 3), (6, 2), (6, 2)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 8=  $\frac{4}{36}= \frac{1}{9}$
case of getting sum 9 = (6, 3), (6, 3)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 9=  $\frac{2}{36}= \frac{1}{18}$