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.

Watch the complete video that explains the solution with 3 different approaches.

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[ashokkhullar]

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.

[raghavxd]

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

[manish]

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

[emem]

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...;)

[gokuruto-yt]

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..

[unknownplayer]

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

[md.afzalbari]

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

[adarshraj]

Nice explanation..
Keep posting awesome questions☺️

[CultOfJ]

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

[schnipsikabel]

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

[nandapeela]

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

[douglasfeather]

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

[atulsharma]

It's the concept of ordering only

[shubhampatil]

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 =

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 = 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.

[DinoCat]

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

[prateekjain]

Nice explanation
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

[ellaenchanted]

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.

[jacoboribilik]

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

[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.

[onlysyn]

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.