4 dices are thrown simultaneously. find the probability that 2 of them show same face and remaining 2 show different faces??

1) 4/9
2) 5/9
3) 11/18
4) 7/9

• ans is (6X6X20)/(6^4)= 5/9
• 11 years agoHelpfull: Yes(1) No(0)
• Total Outcome=6x6x6x6
  
  Favorable 16x6
  
  so 16x6/6x6x6x6=2/27
• 11 years agoHelpfull: Yes(0) No(0)
• Select a number which occurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in , ways.
  Now select two distinct number out of remaining 5 numbers which can be done in ways. Thus these 4 numbers can be arranged in 4!/2! ways.
  So, the number of ways in which two dice show the same face and the remaining two show different faces is
  
  => n(E) = 720
  
• 9 years agoHelpfull: Yes(0) No(0)
• total possible outcomes =6*6*6*6
  now,
  ways in which two positions are selected from 4 is 4C2.
  Now the rest two places are filled as 6!/2!.
  so we get 4C2*6*5*4/6*6*6*6=5/9
• 9 years agoHelpfull: Yes(0) No(0)