two fair dice throws find the probability of the event that the sum of the scores on their upper faces is a perfect square and a prime number
GAURAV1998:
need solutions and answer any?
Answers
Answered by
2
PERFECT SQUARE:1\12
PRIME:5\36
PRIME:5\36
Answered by
4
sample space ={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1) ,.......(2,6),
(3,1),......(3,6),
(4,1),....,(4,6),
(5,1),......(5,6),
(6,1),......(6,6)}
total possible out comes = n= 36
1)sum of the upper faces is a perfect squares ={ (1,3),(2,2),(3,1),(3,6),(4,5),(5,4),(6,3)}
number of favorable out comes = m =7
p(E) = m/n = 7/36
2) sum of the upper faces is prime ={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}
number of favorable out comes = m= 15
p(E) = m/n =15/36 =5/12
(2,1) ,.......(2,6),
(3,1),......(3,6),
(4,1),....,(4,6),
(5,1),......(5,6),
(6,1),......(6,6)}
total possible out comes = n= 36
1)sum of the upper faces is a perfect squares ={ (1,3),(2,2),(3,1),(3,6),(4,5),(5,4),(6,3)}
number of favorable out comes = m =7
p(E) = m/n = 7/36
2) sum of the upper faces is prime ={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}
number of favorable out comes = m= 15
p(E) = m/n =15/36 =5/12
Similar questions