4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number.

• 13/18 as 1-(probability of not getting same no.)=1-((6*5*4*3)/(6*6*6*6))
• 10 years agoHelpfull: Yes(71) No(12)
• The first person rolls a die, and gets a number.
  
  The second person has a 5/6 chance of rolling a different number.
  
  The third person has a 4/6 chance of rolling a number different from those two.
  
  The fourth person has a 3/6 chance of rolling a number different from all three.
  
  The odds that no two will roll the same number is 5/6*4/6*3/6 = 5/18 = .2777...
  
  That's the odds that no two will roll the same number, so the odds that two (or more) of them will roll the same number is 1-5/18 = 13/18 = .7222...
• 10 years agoHelpfull: Yes(65) No(4)
• 1-((6*5*4*3)/6^4)
  =1-10/36=13/18
  
• 10 years agoHelpfull: Yes(9) No(1)
• One way to solve it is by considering the opposite - That on throwing the dice 4 times, you never get a pair.
  
  In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are 5/6, 4/6 and 3/6 respectively.
  
  So on multiplying we get 6/6*5/6*4/6*3/6 = 5/18
  
  On substracting it from 1, we get 1-(5/18) = 13/18
  
  Therefore 13/18 is the correct answer
• 9 years agoHelpfull: Yes(9) No(1)
• Rakesh pls provide sol. for this
• 10 years agoHelpfull: Yes(6) No(4)
• Probability of atleast two people getting same number is
  
  1/6 * 1/6 = 1/36 (Ans)
• 10 years agoHelpfull: Yes(5) No(40)
• The possibility of getting any no. is 1.
  The probability of getting the same no. again is 1/6.
  The probability of getting the same no. 3rd time is 2/6.
  similarly, the probability of getting the same no. 4th time is 3/6.
  so the ans. is (1*1/6*2/6*3/6)=1/36.
• 9 years agoHelpfull: Yes(3) No(6)
• pls...tell the correct answer
  
• 9 years agoHelpfull: Yes(2) No(2)
• Let's first find the probability that all the people get different numbers.
  
  Outcome of each die can be any of the 6 numbers. Therefore,
  Total number of outcomes
  =
  6
  4
  =64
  
  First die can give any of the 6 numbers
  Second die can give any of the remaining 5 numbers
  Third die can give any of the remaining 4 numbers
  Fourth die can give any of the remaining 3 numbers
  Therefore, total favorable outcomes 6×5×4×3
  
  P(all people get different numbers)
  =
  6
  ×
  5
  ×
  4
  ×
  3
  6
  4
  =
  5
  18
  =6×5×4×364=518
  
  P(at least 2 people get the same number)
  =
  1
  −
  5
  18
  =
  13
  18
  =1−518=1318
• 8 years agoHelpfull: Yes(1) No(2)
• 1-((6*5*4*3)/(6^4))
• 8 years agoHelpfull: Yes(1) No(1)
• By using, 1-(probability of not getting same no. by diiferent people)=1-((6*5*4*3)/(6*6*6*6))
  We get the answer as 13/18.
  simple...
• 5 years agoHelpfull: Yes(1) No(0)
• Ans is→15180/(6^6)→.325
• 10 years agoHelpfull: Yes(0) No(1)
• total =6*6*6*probability for none are same=6c1*5c1*4c1
  probability atleast 2 are same is 6*6*6-6*5*4=96
  ans=96
• 8 years agoHelpfull: Yes(0) No(7)
• Given 4 men throw a die and 2 people get the same number.
  
  1) The first die can give any of the 6 numbers.
  
  2) The Second die can give any of the remaining 5 numbers.
  
  3) The third die can give any of the remaining 4 numbers.
  
  4) The fourth die can give any of the remaining 3 numbers.
  
  So, the total possible outcomes will be = 6 * 5 * 4 * 3.
  
  
  Probability of all getting different numbers = 6 * 5 * 4 * 3/6^4
  
                                                                          = 5/18.
  
  The probability of 2 people get the same number = 1 - 5/18
  
                                                                               = 13/18.
• 6 years agoHelpfull: Yes(0) No(0)