A person is having 5 coins in his pocket.IN india coins are of only 3 denominations 1,2 and 5. He takes out coins in his pocket.What is the probablity of getting a exact sum of 20

• ANS is 0.
  since there can be no combination of 5 coins that can sum up to exact 20.
• 10 years agoHelpfull: Yes(37) No(10)
• totally 5 coins be in his pocket.coins only 1,2,5 rupee coins.
  now takes out coins two 5,two 2,one 1 rupee coins.but sum is 5+5+2+2+1=15 rupees.
  so not like this
  now takes out coins three 5,one 2,one 1 rupee coins.but sum is 5+5+5+2+1=18.so not comes sum is 20
  ANS is 0.
  since there can be no combination of 5 coins that can sum up to exact 20.
• 10 years agoHelpfull: Yes(7) No(0)
• rishikesh 3/3^5=1/81 not 1/27
• 10 years agoHelpfull: Yes(3) No(2)
• I think ans is 1/31 . There is no given that person take out exactly all coin from his pocket .Total cases to take out 5 coins are 31 .
  5+ 10+10+5 +1 = 31 . There is one favourable case 5,5,5,5 .
• 10 years agoHelpfull: Yes(3) No(1)
• Total No of possible outcome = 3*3*3*3*3 (81)
  Total suitable combination = 55551,55552,55555(3) (Because question never said he takes out all the five coins)
  Probability = 3/81 = 1/27
• 10 years agoHelpfull: Yes(3) No(4)
• 1/243
  total possibilities 3*3*3*3*3=243
  we cannot make 20 by taking out 1,2,3 or 5 coins even of maximum value 5,
  so there is only 1 possibilties i.e, 5555
• 10 years agoHelpfull: Yes(3) No(9)
• question was 3 denomination of 1,2 or 5 not and
• 10 years agoHelpfull: Yes(2) No(0)
• 1/5
  he takes out few coins in his pocket, here not given that sum of all coins, we can take out how much coins that will give me the sum of 20. so only 1 way, that is...
  we are taking 4 coins(5,5,5,5) out of 5 coins. here 5,5,5,5 is the only one way.
  hence 1/5 is the ans.
• 10 years agoHelpfull: Yes(2) No(3)
• ans will be zero.friends check this site it will clear your doubts.
  http://math.stackexchange.com/questions/1407966/probability-of-getting-a-exact-sum-of-20-given-a-person-has-5-coins-in-his-pocke
• 9 years agoHelpfull: Yes(2) No(0)
• yehi to sochna hah , kaise ....?? 0 is not in d option ,i remember 2 options
  1) 1/7
  2)1/27
• 10 years agoHelpfull: Yes(1) No(2)
• 1/27. 3 combinatons are there. 3/(3^5)
• 10 years agoHelpfull: Yes(1) No(6)
• Combination of coin will be such that 5*3+2*2+1*1=20
  probablity=1/3*3*3*3*3=1/343
  
• 10 years agoHelpfull: Yes(1) No(4)
• abe gawaro exact 20 ka sum chahie or ye nahi bola k pancho coin nikale hai.....
• 9 years agoHelpfull: Yes(1) No(0)
• @rishikesh :can you tell me whuch 3 combination u have taken?
• 10 years agoHelpfull: Yes(0) No(0)
• so,what is the answer anyway ?is it 1/27 ?
  
• 9 years agoHelpfull: Yes(0) No(0)
• Let x,y,zx,y,z be the number of coins of value 1,2,1,2,, and 55, respectively. Since the total is 2020,
  x+2y+5z=20.(1)
  (1)x+2y+5z=20.
  Since there are five coins,
  x+y+z=5.(2)
  (2)x+y+z=5.
  Subtracting four times (2) from the (1),
  (x+2y+5z)−4(x+y+z)z−2y−3xz=0=0=3x+2y
  (x+2y+5z)−4(x+y+z)=0z−2y−3x=0z=3x+2y
  Then (2) becomes
  x+y+(3x+2y)=54x+3y=5
  x+y+(3x+2y)=54x+3y=5
  But it is clearly impossible to make 55 with a nonnegative number of 44s and 33s. In other words, this situation is impossible, so the probability is 00.
• 9 years agoHelpfull: Yes(0) No(0)