Riddles Options

[vivekpm]

grrr.... that is right
C should take first aim in the air.. so that B who has better chances of killing gets to A .. if he succeeds.. good enough.. else A will surely eliminate B. And then C next round main dekhega.. kya karna hai.
ye lo one more as the winning prize

A room contains 50 chairs, numbered 1 through 50. Outside of the room, waiting to come in are 50 people, also numbered 1 through 50. The relevant information is this: a) the people enter the room and take a seat in sequential order and for people 2-50, if their seat is available, they take it, otherwise their selection is random and c) person number 1 chooses his seat randomly.

You're person number 50. What is the probability that you end up in your seat? (as in, seat number 50) Explain your answer.

Me not reading this one now... Want to sleep tight now else I most probably will get dreams of probability

Par ShivaniJi, maine apne pehle post pe bhi to kaha tha, if possible "he should misfire the shot". I think hitting on ground is safe, koi bichare bird ki zindagi kyon khatre mai dalni...

Cheers,

[Mandrake]

Sorry fellas, I am not happy with that solution. You say C's chances are 1 out of 3. That doesn't guarantee that only the THIRD shot finds its mark. ANY 1 of three... what if the first shot was accurate?

[shivani]

mmm..
When it is A's turn he can and will shot C with the probability 1. Therefore C has to maximise the probability that A is shot. If C shoots in the air the probability A is shot is 0.5
If C shoots at B than there is 2/3 probability B survives and shoots at A. In this case A is shot with the probability 2/3 * 0.5 = 1/3.
The best strategy for C is to shoot in the air.

survival probability for C when he shoots in air for first time - (1/3 + 1/2)/2 = 5/12
This number needs to be compared with the Kill A scenario. (Killing B is obviously bad.) If C kills A, C's survival probability is 1- (1/2 + 1/2 * 1/3 + 1/2 * 1/3^2 + 1/2*1/3^3...) = 1/4.
Since 1/4 < 5/12, shooting in the air is correct.

[Mandrake]

So C depends of B's 50:50 chance of killing A.

And I thought you were saying 'all dependence is misery - self dependence is joy'....

so much for....

[shivani]

grrr.... that is right
C should take first aim in the air.. so that B who has better chances of killing gets to A .. if he succeeds.. good enough.. else A will surely eliminate B. And then C next round main dekhega.. kya karna hai.

Par ShivaniJi, maine apne pehle post pe bhi to kaha tha, if possible "he should misfire the shot". I think hitting on ground is safe, koi bichare bird ki zindagi kyon khatre mai dalni...

Cheers,

Maine grrrr isiliye to kiya tha Vivek cs you were right again.. and ni Ji pls..

[shivani]

So C depends of B's 50:50 chance of killing A.

And I thought you were saying 'all dependence is misery - self dependence is joy'....

so much for....

Self dependence doesnt mena that one stops thinking and only try to survive on brute force.. saam daam dand bhed.. fgtn the fifth one

[Mandrake]

Forgotten the fifth one?
Rrrrrreally poor memory...

Coz there is no fifth one

[shivani]

nahi hai? why have I always lived with the idea that there is something else other than these 4 and am forgettign it
why did I think there ar 5 to start with????

[Mandrake]

Woh kya hai na, ki hamaare dimensions mein to 4 hi hai...

Kya jaane, aapke multiple dimensions mein aur bhi honge toh...

[shivani]

Vivekaahh... or anyone else!! no attempts at next one at all??

[vivekpm]

Vivekaahh... or anyone else!! no attempts at next one at all??

Was a bit busy for past couple of days. I have attempted this one, not sure if it is correct though. Will post my attempt tomorrow.

Cheers,

[vivekpm]

There are three scenarios to consider here:

1. Person 1 takes his seat(probability=1/50): In this case, last person will definately get his seat (probability=1)
2. Person 1 takes 50th person's seat(probability=1/50): In this case,last person will definately not get his seat (probability=0)
3. Person 1 takes any other seat apart from 1st and 50th(probability=48/50), in this case probability of last person getting his seat is 0.5 because:
   • Person whose seat is taken may take 50th seat or 1st seat or any other seat. And this will continue until someone either selects 50th seat or 1st seat in which case problem is solved. Probability of selecting one out of 1st and 50th seat is 0.5.

So now, from 1,2 and 3:

1/50*1+1/50*0+48/50*0.5=0.2+0.48=0.5

So the answer is 0.5.

Cheers,

[vivekpm]

Kya Hua? Galat Jawab?

Ok... since you are not around, let me post a question which can be solved using very basic Maths. It is not a riddle but involves playing around with numbers.

How to get 24 using "3" twice and "7" twice and using only +,-,*,/

Cheers,

[divz]

Kya Hua? Galat Jawab?

Ok... since you are not around, let me post a question which can be solved using very basic Maths. It is not a riddle but involves playing around with numbers.

How to get 24 using "3" twice and "7" twice and using only +,-,*,/

Cheers,

ur answer seem to be perfec.. well i dint solve it...

are we supposed to use 3 and 7 only or any other no is allowed???????

[vivekpm]

Kya Hua? Galat Jawab?

Ok... since you are not around, let me post a question which can be solved using very basic Maths. It is not a riddle but involves playing around with numbers.

How to get 24 using "3" twice and "7" twice and using only +,-,*,/

Cheers,

ur answer seem to be perfec.. well i dint solved it...

are we supposed to use 3 and 7 only or any other no is allowed???????

Yes you are supposed to use only 3 and 7. To be precise you need to use 3,3,7,7. Nothing more, nothing less.

Cheers,