Two dices are thrown. What is the probality that the the number on the first dice is greater or equal to number on the second dice?

• If first number is 6, then the probality that the the number on the first dice is greater or equal to number on the second dice = 1/6 *6/6 = 6/36=1/6
  If first number is 5, then the probality that the the number on the first dice is greater or equal to number on the second dice = 1/6 * 5/6 = 5/36
  If first number is 4, then the probality that the the number on the first dice is greater or equal to number on the second dice = 1/6 *4/6=4/36
  ..
  ,,
  considering all cases, the probality that the the number on the first dice is greater or equal to number on the second dice = 6/36 +5/36 +4/36 +..+1/36 =
  (1+2+3+4+5+6)/36 = 21/36 = 7/12
• 12 years agoHelpfull: Yes(3) No(0)
• Total outcomes will be 6*6=36 like (1,1)(1,2),......,(6,6)
  Out of that total possible conditions=21 like(1,1)(2,1)(2,2)(3,1)(3,2)(3,3)(4,1)(4,2)(4,3)(4,4)(5,1)(5,2)(5,3)(5,4)(5,5)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
  Therefore Probability=21/36=7/12
• 12 years agoHelpfull: Yes(3) No(3)
• 1/6*1/6+1/6*2/6+1/6*3/6+1/6*4/6+1/6*5/6+1/6*6/6=7/12.........in every multiplication term 1/6 stands for probability of getting 1,2,3,4,5,6 on 1st die respectively and 2nd term for getting ......ex.if we get 3 on 1st die then we can have 1,2,3 on 2nd die thus 1/6*3/6....similarly for all cases
• 12 years agoHelpfull: Yes(0) No(1)