Brilliant Logical Probability Puzzle || A, B, C Speaking with 5 other persons

3 persons A,B,C need to speak at a function with 5 other persons. These 8 Persons need to speak in random order.
Whats the probability that A speaks before B and B speaks before C.
So we are interested only in the order of A B and C.... we don't care about the order of other 5 persons.
Im thankful to Mr. Babu Ram who shared me this on facebook.
It was asked to him in an interview at a software organization.
I highly encourage all of you to share as many questions as you can on my gmail or facebook and contribute to this logical community.
Gmail : [e-mail]
Facebook : / mohammmedammar
Watch the complete video that explains the solution with 3 different approaches.
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[@ashokkhullar6650]

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.

[@blackdalia3912]

exactly what i thought

[@manish2359]

It is one of the easiest puzzles you shared, solved it in 3-4 seconds and logically not mathematically.

[@raghavxd3161]

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

[@BeholdingKrishna]

I did the same

[@raismin739]

me too, the first touth was the other persons don't matter, but i didn't give a reason why

[@Havefunatwork]

Buncha nerds (geniuses)

[@wernerviehhauser94]

I split them in two groups and imagined I only watched when one of the (A,B,C) groups speaks. This yields the correct result, but fails to prove its correctness.

[@N8570E]

Exactly. Well done. And nicely explained. The irrelevant is irrelevant.

[@emem2756]

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...;)

[@md.afzalbari4890]

I always like mathematical approach... bec I am maths lover..

[@adarshraj5167]

Nice explanation.. Keep posting awesome questions☺️

[@ClickBeetleTV]

This one was great! Thanks!

[@aasawari07]

Explaination 🙌🏻🙌🏻🙌🏻

[@CultOfJ]

Well done 👍😊 Thnx for posting such videos👌Keep posting new LOGICAL videos every week..!! 🔔This video was awesome..!!😎

[@LOGICALLYYOURS]

Thanks bro :)

[@DinoCat259]

I done it just after reading the question ihave solved lots of problems of permutations and combinations

[@nandapeela1080]

Your videos are simply mind blowing.... Amazing...

[@LOGICALLYYOURS]

Thanks Nanda :)

[@atulsharma8067]

It's the concept of ordering only

[@sunitdas6282]

Nice

[@schnipsikabel]

I like your riddles a lot , but could you please emphasize 'probability' on the 3. syllable? Thanks ;)

[@gokuruto-yt3688]

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..

[@unknownplayer6988]

By P and C possible outcome 8p3/3 and total outcome is 8! So probability is 1/6

[@aryanshrajsaxena6961]

Solved it mentally 🦾

[@ellaenchanted2399]

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.

[@alexandershishlev7428]

I had a bit other impression and solved it with given A is first, then B, then C, then anyone else. A has a chance of 1/8 to speak first, then B has a chance 1/7, as A already spoke and out of line, the same for C, 1/6. So, the result is (1/8)*(1/7)*(1/6)=(1/336). In your case, ABC in any place but in a group, consider this, if there was only X, total number of combinations would be 4! or 24, but only 2 we are looking for XABC or ABCX, (2/24 or 1/12) with 5 people it's 5! or 120 combinations, but we are looking for 6 (ABCXY, ABCYX, XABCY, YABCX, XYABC, YXABC) which will be 6/120 or 1/20. Did you notice the number of combinations we are looking for is (1+n)! where n is a number or additional elements? That's because we consider ABC group as one element. So, the function will be (1+n)!/m! where n is a number of additional members, and m is a total number of members, and the result is (1+5)!/8!=>6!/8!=>1/(7*8)=>1/56.

[@saurabhkawatra8938]

The answer to your interpretation would be 6c1*5!/8! i.e 1/56

[@sumandey3934]

Babu ram .. Konse company k questn hai ?

[@ShivamMishra-td3jz]

Ammar sir rocks😎

[@LOGICALLYYOURS]

Shivam... thanks bro for the appreciation :)

[@jacoboribilik3253]

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

[@douglasfeather3745]

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

[@atulsharma8067]

It's very simple 1/6

[@desidoodperfect]

Waiting!!?

[@nyanmaru582]

Wtf..! it was hard af.!

[@davidbornstein9197]

My way was easier than any of these! The odds A goes before the other 2 is 1/3 and the odds that after this B goes before C is 1/2. Multiply those two numbers to get the answer.

[@cipherxen2]

If order doesn't matter, ignore them.

[@shubhampatil155]

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.

[@aeSaaket]

Wow nice where will u gett these ideas from

[@sandeepmeena5932]

From cengege book

[@grzeskowalski]

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.

[@atulsharma8067]

I m a mathematics teacher and atleast 20 students of my class can give the answer in 10 sec

[@jacoboribilik3253]

so you think they are geniuses or this is easy peasy?

[@onlysyn1]

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.

[@sudheerthunga2155]

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..

[@LOGICALLYYOURS]

Sudheer.. the point is... 8P3 means all possible ways A,B,C can be arranged... but the number of valid ways (where A speaks before B and B speaks before C) can be calculated using 8C3. For simplicity... consider, there are only three persons A,B,C... possible ways of arrangements : 3P3 (i.e. 6)..... but number of arrangements in which they speak in a particular order (A->B->C) : 3C3 (i.e. 1 and that's nothing but ABC)... so probaility is : 3C3 / 3! = 1/6.

[@Grassmpl]

I have an easier explanation. Think of positions that each person is standing as 1,2,3,...8. Where 1 mean first and 8 means last. Thus we must associate each of the 8 people with a different integer between 1 and 8. A,B,C are 3 PARTICULAR people (eg Alan,Bob,Chris). They need to be associated with numbers a

[@prateekjain2473]

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

[@LOGICALLYYOURS]

Thanks Prateek... There are two types of defective coins questions.. 1 - 10 bags of coins, one bag has all defective coins, use weighing machine only once... I have posted this puzzle : kzbin.info/www/bejne/bJKbpnpvgphgmNE 2 - Other type is where you have several coins and using weighing machine minimum number of times you need to find defective coin. Your question seems to be the first one, but you haven't mentioned BAGS. Please check out the video that I mentioned above "10 Bags Puzzle || Find the bag with defective coins" and let me know if it's similar question.

[@parveengoel9051]

The soln. is not possible

[@prateekjain2473]

Interviewer only asked about coins he doesn't mentioned bags...That's why I was confused during interview....

[@Roshan20107]

1. Put 5 coins in each side of the scale 2. One side should be lighter and has the defective coin 3. Take one coin from each side, see if the balance becomes straight, if it becomes straight, then the coin taken from lighter side is defective 4. If balance doesn't become straight, take another coin from each side. Keep repeating till you find the defective coin

[@harshrajsharma5050]

Put all the the 10 coins on the weighing machine and remove one by one a coin ,the coin which will make a difference of 4g in total weight is the defective coin

[@OceanAce]

(Probability of A not B not C) x (Probability A then B) ?? I hate these type of problems

[@TheDigiWorld]

I took a very complicated mathematics approach, which is impossible to explain in comments

[@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?

[@CultOfJ]

Assume that each balls weighs 3gms and the ball having diamond weighs 4gms and general formula for weighing machine is: LEFT-(minus)RIGHT. Now let's take the 5 pairs as A, B, C, D, E and without loss of generality we assume that E has the diamonds. Now put each ball from A,B and C on the Left pan and each from D and E on the Right one. Left should weigh 3+3+3=9 while Right should weigh 3+3=6 (since we don't know in which ball the diamonds are). If the digital display show +2 means one of the balls in Right pan has a diamond OR if it shows +4 means one of the three balls in Left pan has the diamond. Case 1: Ball is in Right pan. Simply weigh between two balls one on the either side. If display shows +1 means The Left one has diamond and Of -1 means The Right one. Case 2: The Left pan has that ball. Put any two balls on the Left and the remaining on the right. Now if it shows +4 means any one ball of the Left pan has diamond, you can find that by following the same method as in CASE 1. If it shows +2 means The ball in the Right pan has that diamond. But at first you have to find weight of each ball by comparing any 2 balls of different colour. Hope you understood. Thnx for asking..!!😉😎👍

[@geostanyjose]

The answer is 1. We need to use the weighing pan only once. Assume that all the balls weigh 5 gm and the pair of balls with the diamond weighs 6 gram. We can name the pairs as A, B, C, D and E. The formula is Left pan- (minus)Right pan In the left pan put 1 A colour ball and 2 B colour balls. In the right pan put 1 C ball and 2 D colour balls. Here are the five different possibilities that will show in the weighing scale. 0 : This means diamond is in the E colour balls which we were not used. 1 : The diamond is in the A colour balls 2 : The diamond is in the B colour balls -1: The diamond is in the C colour balls -2: The diamond is in the D colour balls

[@CultOfJ]

@@geostanyjose Ohh great bro..!! It's short and precise. Great logic..!!

[@disguisedhell]

@@geostanyjose yes, that's exactly how I made it

[@Havefunatwork]

Problabilty😁

[@harisali5025]

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

[@DanielWalvin]

If I have to hear "proBABble-ity" again, I might throw my phone out of a window.

[@cjfdnqkn4374]

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

[@desidoodperfect]

Face reveal

[@LOGICALLYYOURS]

almost there.... just 4 kgs away :) hope you got my point ;)

[@desidoodperfect]

@@LOGICALLYYOURS ok last bhana h yeh sir pllss

[@LOGICALLYYOURS]

yes.. pakka ye last one :)

[@jouher5747]

100k

[@naomipasirpanjang7045]

Mistake at 3:00 there sre 2 &

[@atulsharma8067]

No need to apply such a big calculation

[@user-rl5tg1uy4m]

Why do u care about the other 5 lol. 1/3!=1/6

[@nikolabyalmarkov3582]

I watched the video carefully and I clearly see that in the beginning you asked one question - "What is the probability A speaks before B, and B speaks before C?", but then suddenly you decide to answer to a totally different question ( more like - how many variations exist and how many of them are in the term of the question). But actually the original question, which you asked in the beginning has only one answer and is very simple : Probability is exactly 50%, because or this event will happen (A->B->C, no matter if there are any other speaker between them), or this event will not happen (50 %), because in the real life or some certain event happens or not. I am agree that your unnecessary complications are correct, but only if ask another question, so anyway it is interesting to see how someone is asking one simple question and then answer to another with a lot of complications. In the end of the story we can learn from every one different experience, for good or bad. Knowledge is the best.

[@michaelkhouzam]

Let me ask you this Ammar, what is that proBabbillity that you are pronouncing the word probability in a very funny way?