Three dice are rolled simultaneously. What are the number of possible outcomes such that at least one shows the number3?
option
A. 63
B. 108
C. None
D. 81
E. 91

• jst use concept of probability and permutation and combination.
  the no. of possible ways of not having 3 in any of the dice is = 5c1*5c1*5c*1=125
  and total no possible ways is = 6c1*6c1*6c1=216
  the no of ways having atleast one 3 is = 216-125=91
  
• 11 years agoHelpfull: Yes(12) No(0)
• no of outcomes in which 3 is not showing= 125
  no in which 3 is included= (216-125)= 91
• 11 years agoHelpfull: Yes(3) No(0)
• =1-(5/6*5/6*5/6)
  =1-125/216
  =91
• 10 years agoHelpfull: Yes(2) No(0)
• ans will be none of this
  explanation:
  three dice are rolled so possible outcomes of atleast one shows number 3
  1st 2nd 3rd
  1 . 0 0 3
  2 . 0 3 0
  3 . 3 0 0
  4 . 0 3 3
  5 . 3 0 3
  6 . 3 3 0
  7 . 3 3 3
  
  so ans will be 7
• 11 years agoHelpfull: Yes(0) No(12)
• ans is 108
• 11 years agoHelpfull: Yes(0) No(2)
• b will be the answer
• 11 years agoHelpfull: Yes(0) No(0)
• ans is 91
  
• 9 years agoHelpfull: Yes(0) No(0)