Latest Elitmus Aptitude Question SOLUTION: 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

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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

Asked In Elitmus harsh (9 years ago)

Unsolved Read Solution (16)

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Rishikesh Sharan  (9 years ago)

ANS is 0.
since there can be no combination of 5 coins that can sum up to exact 20.

Is this solution Helpfull? Yes (37) | No (10)  

Mani kandan  (9 years ago)

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.

Is this solution Helpfull? Yes (7) | No   

Prateek Pandey  (9 years ago)

rishikesh 3/3^5=1/81 not 1/27

Is this solution Helpfull? Yes (3) | No (2)  

kapil  (9 years ago)

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 .

Is this solution Helpfull? Yes (3) | No (1)  

Deepesh Samant  (9 years ago)

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

Is this solution Helpfull? Yes (3) | No (4)  

HARSH AGRAWAL  (9 years ago)

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

Is this solution Helpfull? Yes (3) | No (9)  

sudarshan  (9 years ago)

question was 3 denomination of 1,2 or 5 not and

Is this solution Helpfull? Yes (2) | No   

subhash kumar  (9 years ago)

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.

Is this solution Helpfull? Yes (2) | No (3)  

swarup  (8 years ago)

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

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harsh  (9 years ago)

yehi to sochna hah , kaise ....?? 0 is not in d option ,i remember 2 options
1) 1/7
2)1/27

Is this solution Helpfull? Yes (1) | No (2)  

Rishikesh Sharan  (9 years ago)

1/27. 3 combinatons are there. 3/(3^5)

Is this solution Helpfull? Yes (1) | No (6)  

Pooja bala singh  (9 years ago)

Combination of coin will be such that 5*3+2*2+1*1=20
probablity=1/3*3*3*3*3=1/343

Is this solution Helpfull? Yes (1) | No (4)  

Arpit dubey  (8 years ago)

abe gawaro exact 20 ka sum chahie or ye nahi bola k pancho coin nikale hai.....

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Prakash Singh  (9 years ago)

@rishikesh :can you tell me whuch 3 combination u have taken?

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Agniv Chatterjee  (8 years ago)

so,what is the answer anyway ?is it 1/27 ?

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lokesh kumar  (8 years ago)

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.

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