Four dice are rolled simultaneously. What is the number of possible outcomes in which at least one of the die shows 6?

• there are 4 dies so total outcomes are 6^4=1296
  the no of outcomes in which none of the dies show 6 will be 5^4=625
  now no of outcomes in which at least one die show 6=>1296-625=671
  
• 13 years agoHelpfull: Yes(69) No(3)
• When 4 dice are rolled simultaneously, there are 6^4 = 1296 outcomes.
  The converse of what is asked in the question is that none of the dice show '6'. That is all four dice show any of the other 5 numbers.
  That is possible in 5^4 = 625 outcomes.
  Therefore, in 1296 - 625 = 671 outcomes at least one of the dice will show 6..
• 13 years agoHelpfull: Yes(19) No(3)
• Probability of getting 6 in one roll= 1/6 ;
  Probability of not getting 6 in one roll= 1-1/6 = 5/6;
  Now,
  For 4 DICE
  Probability of not getting 6 in "FOUR" roll= (5/6)^4 = 625/1296;
  So, In total 1296 times we will not get a 6 in 625 times, that means in rest of all in 1296 (ie, 1296-625) we will get at least one 6.
  So, Answer is = (1296-625) / 1296 = 671/1296.
  
• 10 years agoHelpfull: Yes(6) No(3)
• one die to show 6, probability is 4 times.
  two dies to show 6, probability is 3 times.
  three to show 6, probability is 2 times.
  four dies to show 6, probability is 1 time.
  Hence 4*3*2*1=36
• 13 years agoHelpfull: Yes(4) No(36)
• here i question rearrange so ans is 4!-1= 23
  
• 11 years agoHelpfull: Yes(2) No(6)