@Torque024 said:color diff hai lakin ismai, confused :/
bead toh symetric hota hai na bhai ?
@Torque024 said:Ram is attending a disco party. Upon arriving, each of the K party members including ram puts his or her coat on the coatrack. After party, the first person to leave grabs a coat at random from the coatrack. As the rest of the members leave, they try to grab their own coat, but if their coat has already been taken, they take someone else's coat at random from the rack. Ram is the last person to leave the party. What is the probability that ram will go home with his own coat?PS: Don't have OA for this.
Leaving the case of Ram,
the rest of people will be arranged among themself with respect to their coat in (k-1)! ways
so no of favorable cases: [(k-1)!]
total case: k!
probability: [(k-1)!]/k!= 1/k
Correct me if am wrong
@Zedai said:[(1/k)-(1/k!)]Leaving the case of Ram, the rest of people will be arranged among themself with respect to their coat in (k-1)! ways out of which in one case they all will be arranged properly (i.e each one will get his own coat)so no of favorable cases: [(k-1)!-1]total case: k!probability: [(k-1)!-1]/k!= [(1/k)-(1/k!)]Correct me if am wrong
Why [(k-1)!-1]? I mean Why the -1 part?
If they are arranged properly... Its not violating any condition....
If they are arranged properly... Its not violating any condition....
@adwaitjw said:Why [(k-1)!-1]? I mean Why the -1 part?If they are arranged properly... Its not violating any condition....
they try to grab their own coat, but if their coat has already been taken, they take someone else's coat at random from the rack
missed the bold area! 
@pankaj1988 said:1/2...This is same as ipod ques of one of the AIMCATS..
OA is 1/2
Explain the logic.
Explain the logic.
@Torque024 said:Ram is attending a disco party. Upon arriving, each of the K party members including ram puts his or her coat on the coatrack. After party, the first person to leave grabs a coat at random from the coatrack. As the rest of the members leave, they try to grab their own coat, but if their coat has already been taken, they take someone else's coat at random from the rack. Ram is the last person to leave the party. What is the probability that ram will go home with his own coat?
got the logic now..
read the question wrong..
it will be 1/2.
@Zedai said:got the logic now..read the question wrong..it will be 1/2.
What's the prob with 1/k .. what's wrong in that approach?
@Torque024 said:OA is 1/2 Explain the logic.
@adwaitjw said:Can u explain how?
oh....muje laga wrong hai isliye maine delete kar diya..
Problem is more of a logic than calculation... ...ans will always be 1/2 irrespective of number of people...Like in AIMCAT there were 4 people...
Suppose any person X picks the coat of the last person(Ram) than all the subsequent persons will get their own coat and Ram will get the coat of X..
So the probability of Ram getting his own coat= Person X getting his own coat..(either he get on don't get)
P=1/2
P.S. In problem its written that all the persons try to get their own coat...
@pankaj1988 said:oh....muje laga wrong hai isliye maine delete kar diya..Problem is more of a logic than calculation... ...ans will always be 1/2 irrespective of number of people...Like in AIMCAT there were 4 people...Suppose any person X picks the coat of the last person(Ram) than all the subsequent persons will get their own coat and Ram will get the coat of X..So the probability of Ram getting his own coat= Person X getting his own coat..(either he get on don't get)P=1/2P.S. In problem its written that all the persons try to get their own coat...
@Torque024 said:Ram is attending a disco party. Upon arriving, each of the K party members including ram puts his or her coat on the coatrack. After party, the first person to leave grabs a coat at random from the coatrack. As the rest of the members leave, they try to grab their own coat, but if their coat has already been taken, they take someone else's coat at random from the rack. Ram is the last person to leave the party. What is the probability that ram will go home with his own coat?
1/k!?
It's 1/2.@Torque024 said:It's 1/2.
@Torque024 @deedeedudu bhai.. iska logic ye hai ki suppose there are 4 people... and 1st picked up the coat of 3rd person, so now 3 coats are left... as soon as someone picks up the coat of 1st person, everone gets his/her coat in this scenario... so logic is either the person gets his coat or not.. hence 1/2... yahin pe PG pe explain hua tha ye.. aise cases me answer hamesha 1/2 rahega
If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
@grkkrg said:If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
2,2,1-> 5c2*3c2*2
3,1,1-> 5c3*2c1*2
total = 100 ??
total = 100 ??
@adwaitjw said:Can u explain how?
@htomar said:What's the prob with 1/k .. what's wrong in that approach?
see, the logic here is the last person will either get his own ipod or the ipod of the last person.
the probability of the picking the ipod of the 1st person is equal to the prob of picking the ipod of the last person
here's how:
if you think over the arrangement closely enough, you will observe that whenever a person picks up the ipod of the 1st person... thereafter everyone gets their own ipod...
if the 1st person gets his ipod , everyone will get their ipod.
and if he misses his ipod, the rest of the people will get their ipod only from the time a person picks up the ipod of the 1st person.
So, the last person gets his own ipod or the ipod of the 1st person.
therefore prob: 1/2
@grkkrg said:If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
Same as distributing 5 different objects in 3 distinct boxes, with at least one object in each
Case 1: 2 2 1 => (5!/2!2!1!2!)*3! = 90
Case 2: 3 1 1 => (5!/3!1!1!2!)*3! = 60
Total = 150 onto functions
@grkkrg said:If A=(p,q,r,s,t) and B=(x,y,z), how many onto functions A->B exist?
(-1)^2*(1)^5*3c1+(-1)^1*(2)^5*3c2+(-1)^0*(3)^5*3c3
=3-96+243
=150
=3-96+243
=150