a student rolls two 6 sided red die and one 6 sided blue die what is the probability that number shown on the blue die exceeds both the number on the red die a)1/6 b)5/36 c)55/216 d)1/2

• Let the blue die rolls to 6, so the combinations available are
  6>(1,1)
  >(1,2)
  >(1,3)
  .
  .
  .
  >(5,5)
  So a total of 5*5=25 combinations,
  similar case if the blue die rolls to 5 we will have 4*4=16 combinations, similar for all the other blue dice cases...hence the total is 5*5+4*4+3*3+2*2+1=55
  thus the probability 55/216
  Thus the t
• 8 years agoHelpfull: Yes(10) No(0)
• Someone please xplntn the method clearly....
  
• 8 years agoHelpfull: Yes(2) No(0)
• solution is : 55/216
• 8 years agoHelpfull: Yes(1) No(1)
• C) 55/216
• 8 years agoHelpfull: Yes(1) No(0)
• B 5/36
  2dies are rolled so total no of outcomes 36
  The no can't be 7 in a single die . coz 6 face .
  So it can be 2,3,4,5,6, total 5no
  So it ts 5 /36
• 8 years agoHelpfull: Yes(0) No(5)
• blue die must be 3,4,5,6 cases (2 can't be possible because 2 red die atleast 1 and 1 means sum =2 and in given question blue die must be greater than sum of two red dies number ok)
  
  1. sum 3==(1,1)=1
  2.sum 4=(1,1),(1,2),(2,1)=3
  3.sum 5=(1,1),(1,2),(2,1),(2,2),(3,1),(1,3)=6
  4.sum 6=(1,1),(1,2),(2,1),(2,2),(3,1),(1,3),(1,4),(4,1),(5,1),(1,5)=10
  
  p(blue
• 8 years agoHelpfull: Yes(0) No(2)