Brilliant Logical Probability Puzzle A, B, C Speaking with 5 other persons
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Date: 2023-11-15
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Comments and reviews: 20
disguisedhell
Try this one
Suppose we are given 5 pairs of balls in 5 colours, such that all the balls have the same weight.
Now we are told that someone has put two 1 gram diamonds in one of the 5 pairs of balls, so that now there is exactly one pair of balls having same colour, where the two balls each are 1 gram heavier than the other balls.
You are given a scale with two pans, marked as the 'left pan' and the 'right pan', and a digital display which shows the signed difference in the weights (in grams) between the 'left pan' and the 'right pan'.
So if we put 3 grams in the 'left pan' and 5 grams in the 'right pan' it will show -2 grams, and if we swap the weights it will show +2 grams.
You can put as many balls into each of the pan as you like.
What is the minimum number of weighings that you need to take in order to determine the colour of the balls containing the diamonds?
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Try this one
Suppose we are given 5 pairs of balls in 5 colours, such that all the balls have the same weight.
Now we are told that someone has put two 1 gram diamonds in one of the 5 pairs of balls, so that now there is exactly one pair of balls having same colour, where the two balls each are 1 gram heavier than the other balls.
You are given a scale with two pans, marked as the 'left pan' and the 'right pan', and a digital display which shows the signed difference in the weights (in grams) between the 'left pan' and the 'right pan'.
So if we put 3 grams in the 'left pan' and 5 grams in the 'right pan' it will show -2 grams, and if we swap the weights it will show +2 grams.
You can put as many balls into each of the pan as you like.
What is the minimum number of weighings that you need to take in order to determine the colour of the balls containing the diamonds?
reply
shubham
All the solution provided are so mathematical and complicated. I don't think a normal person who only knows basic maths won't understand it. So. My solution is. Calculating chances. If we consider 3 people only. As only 3 matter. What is the chance of A getting to speak before B, and C. Its 1 in 3. If we consider percentage
Its 100/3 = 33. 33333
Now that we got A covered lets just focus on B And C. What is the chance that B gets to speak before C its 1 in 2
That is Half. So. Out of all the times A gets to speak first. Only half the times B will get to speak second. So its half the times of the chances of A being first
That is 33. 33333/2 = 16. 66
So there is a 16. 66% chance that A will speak before B and B will speak before C which is exactly 1/6 probablity.
reply
All the solution provided are so mathematical and complicated. I don't think a normal person who only knows basic maths won't understand it. So. My solution is. Calculating chances. If we consider 3 people only. As only 3 matter. What is the chance of A getting to speak before B, and C. Its 1 in 3. If we consider percentage
Its 100/3 = 33. 33333
Now that we got A covered lets just focus on B And C. What is the chance that B gets to speak before C its 1 in 2
That is Half. So. Out of all the times A gets to speak first. Only half the times B will get to speak second. So its half the times of the chances of A being first
That is 33. 33333/2 = 16. 66
So there is a 16. 66% chance that A will speak before B and B will speak before C which is exactly 1/6 probablity.
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akash
Hi,
I didn't understand mathematical solution of your's, which is (8C3 x 5! ) / 8!
I think, it is very easy to understaned the following way.
Let's first allow the other 5 person to sit, it is like 8 place and 5 persion (8P5, after this there is only one way ABC can sit. .ie.
X - X X X X - -; X is occupied and - will be taken by ABC, now there is only one way ABC can speak.
final result is the same. But I feel this is better to understand.
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Hi,
I didn't understand mathematical solution of your's, which is (8C3 x 5! ) / 8!
I think, it is very easy to understaned the following way.
Let's first allow the other 5 person to sit, it is like 8 place and 5 persion (8P5, after this there is only one way ABC can sit. .ie.
X - X X X X - -; X is occupied and - will be taken by ABC, now there is only one way ABC can speak.
final result is the same. But I feel this is better to understand.
reply
Grzegorz
You place mr B on any position between 2 and 6 (so six times with 1/8 probability) and then, for each position of mr B, you place mr A and then mr C on any place that meets the conditions. That would be:
1/8 - (1/7 + 2/7-5/6 + 3/7-4/6 + 4/7-3/6 + 5/7-2/6 + 6/7-1/6) = 1/6
First 1/7 in the bracket means that we do not need to place mr C, because the assumed placement of mr B and mr A already meets conditions.
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You place mr B on any position between 2 and 6 (so six times with 1/8 probability) and then, for each position of mr B, you place mr A and then mr C on any place that meets the conditions. That would be:
1/8 - (1/7 + 2/7-5/6 + 3/7-4/6 + 4/7-3/6 + 5/7-2/6 + 6/7-1/6) = 1/6
First 1/7 in the bracket means that we do not need to place mr C, because the assumed placement of mr B and mr A already meets conditions.
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GOKURUTO
I solved it with a different approach.
First just neglect the 5 person's because there position doesn't even matter.
Now we have three people A, B and C.
The probability of picking A is 1/3 and after picking A the probability of picking B is 1/2.
Therefore the probability will be 1/3 - 1/2 = 1/6.
reply
I solved it with a different approach.
First just neglect the 5 person's because there position doesn't even matter.
Now we have three people A, B and C.
The probability of picking A is 1/3 and after picking A the probability of picking B is 1/2.
Therefore the probability will be 1/3 - 1/2 = 1/6.
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Juan
Actually if order mattered (the 5 remaining people) it would still be 1/6. You have 5 people and 3 sticks separating the three bunches. How many possible rearrengements are possible: 8! /3! and we have 6 possible ramifications of each out of which one is the desired event==> Prob=1/6
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Actually if order mattered (the 5 remaining people) it would still be 1/6. You have 5 people and 3 sticks separating the three bunches. How many possible rearrengements are possible: 8! /3! and we have 6 possible ramifications of each out of which one is the desired event==> Prob=1/6
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Raghav
I actually got it right in about 5-7 seconds but I did not do it that way I just ignored the 5 people bcz their positions does not matter so we had 3 people
So possible outcomes were 3! i. e 6 and favourable was only 1
Therefore, the answer is 1/6
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I actually got it right in about 5-7 seconds but I did not do it that way I just ignored the 5 people bcz their positions does not matter so we had 3 people
So possible outcomes were 3! i. e 6 and favourable was only 1
Therefore, the answer is 1/6
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prateek
Nice explanation sir.
If there are 10 coins each measure 5 gm. but one coins weights 4 gm. How can we find defective coin because we have to use weigtining machine only once?
This puzzle was asked me during interview
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Nice explanation sir.
If there are 10 coins each measure 5 gm. but one coins weights 4 gm. How can we find defective coin because we have to use weigtining machine only once?
This puzzle was asked me during interview
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Ella
When I first heard this, I interpreted it as A is right before B who is right before C (i. e. ABC(5random people, 1 person ABC 4 others etc etc. I'm curious as to how the probability would shift given this or if it would.
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When I first heard this, I interpreted it as A is right before B who is right before C (i. e. ABC(5random people, 1 person ABC 4 others etc etc. I'm curious as to how the probability would shift given this or if it would.
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Douglas
Agree the answer is stated in the video. However, what is wrong with this approach. P(A before B) = 1/2 and P(B before C) = 1/2. So P(A before B and B before C) = P(A before B) - P(B before C) = 1/2 - 1/2 = 1/4
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Agree the answer is stated in the video. However, what is wrong with this approach. P(A before B) = 1/2 and P(B before C) = 1/2. So P(A before B and B before C) = P(A before B) - P(B before C) = 1/2 - 1/2 = 1/4
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Sudheer
Sir but in your last approach I don't get why it is 8C3 because as far as I know that in our case 8C3 would include the cases which are not valid that is against our question.
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Sir but in your last approach I don't get why it is 8C3 because as far as I know that in our case 8C3 would include the cases which are not valid that is against our question.
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Cjfdn
Think just A, B, and C
There are 6 possible combinations, and the only the ABC combination will be right to the rules.
Therefore, the probability is 1 out of 6, or 1/6
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Think just A, B, and C
There are 6 possible combinations, and the only the ABC combination will be right to the rules.
Therefore, the probability is 1 out of 6, or 1/6
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Ashok
Unnecessarily made to look harder. It's so simple.
A, B, C can be arranged in six ways out of which only one is favourable. So the answer is 1/6.
I solved orally.
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Unnecessarily made to look harder. It's so simple.
A, B, C can be arranged in six ways out of which only one is favourable. So the answer is 1/6.
I solved orally.
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Haris
Just make the samole space considering A, BAnd C only
ABC
ACB
BAC
BCA
CAB
CBA.
P= Required no of outcomes / total no of outcome
P=1/6
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Just make the samole space considering A, BAnd C only
ABC
ACB
BAC
BCA
CAB
CBA.
P= Required no of outcomes / total no of outcome
P=1/6
reply
MyKazam
Let me ask you this Ammar, what is that proBabbillity that you are pronouncing the word probability in a very funny way?
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Let me ask you this Ammar, what is that proBabbillity that you are pronouncing the word probability in a very funny way?
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Cult
Well done -- Thnx for posting such videos-Keep posting new LOGICAL videos every week! -This video was awesome! -
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Well done -- Thnx for posting such videos-Keep posting new LOGICAL videos every week! -This video was awesome! -
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schnipsikabel
I like your riddles a lot, but could you please emphasize 'probability' on the 3. syllable? Thanks; )
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I like your riddles a lot, but could you please emphasize 'probability' on the 3. syllable? Thanks; )
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manish2359
It is one of the easiest puzzles you shared, solved it in 3-4 seconds and logically not mathematically.
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It is one of the easiest puzzles you shared, solved it in 3-4 seconds and logically not mathematically.
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DinoCat
I done it just after reading the question ihave solved lots of problems of permutations and combinations
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I done it just after reading the question ihave solved lots of problems of permutations and combinations
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EM
That-s actually easy: A can be either before or after B, so 50/50 no matter how many there are x, y or z. ;)
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That-s actually easy: A can be either before or after B, so 50/50 no matter how many there are x, y or z. ;)
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