Ram has two dies which has a nos 2,3,4,6 when he throws the two dies together find the no of possibility so that the some of the two dies should be even...???

• total possible combinations=(2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(3,3)(6,2)(6,4)(6,6)=10
  so probability=10/16=5/8
  
• 9 years agoHelpfull: Yes(14) No(0)
• total outcome will be 16, and to get even number it should be a combination of (2,2)(2,4)(2,6). and so on
  counting to 10
  therefore 10/16=5/8
• 9 years agoHelpfull: Yes(1) No(0)
• First can have 4 outcomes
  Second can also have 4 outcomes
  So total combination should be 4×4=16.
  Now, even+even=even and odd+odd=even
  So 3 even numbers . Outcome of even number one on dice is 3. Same in the second one
  So total possible outcome is 3×3=9
  Now last digit i.e odd is 3. As combination of 3 can be with 3 only... as odd+odd=even
  So finally 9+1=10 is total number of outcome for even.
  So it would be 10/16 or 5/8 (possible outcome/total outcome).
• 9 years agoHelpfull: Yes(1) No(0)
• n(e)=(2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(3,3)(6,2)(6,4)(6,6)=10 ,p(e)=16
  probability=n(e)/P(e)=10/16=5/8
• 9 years agoHelpfull: Yes(1) No(0)
• We know ,
  even+even=even or odd +odd =odd
  so out of given numbers(2,3,4,6)...there are only 3 even numbers ..so possibility of getting even number when two dies are thrown is [(2,4) ,(2,6),(4,2) ,(4,6) ,(6,4),(6,2)].....
  
  ....so total no of ways to get even number is 6.
  and total nof of possible outcome is 4*4=16
  so probability of this event is =6/16 i.e 3/8
  
  i hope this answer is helpful
• 9 years agoHelpfull: Yes(1) No(5)
• 5/8 will be the answer apart from the confusion that what no's does the other two #faces posses....
• 9 years agoHelpfull: Yes(1) No(0)
• answer will be 11/18
  bcoz one cube may also show the number, whether other cube may not show any number when rolled
• 9 years agoHelpfull: Yes(0) No(0)