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.

Answers (1)

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}






 



 


 

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