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Probability question
Quantitative Questions and Answers Discuss Quantitative and other Math related questions. Post your math doubts and get it solved by the smartest brains this side of the universe !

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zion1729 zion1729 is offline
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Re: Probability question - 12-06-2008, 06:44 PM

The number of ways of arranging the words of DAUGHTER so that none of the vowels occur together is:

8! - 3!6! - 3.2.(7!-2.6!) = 3.7! - 6! = 14400.


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Re: Probability question - 07-08-2008, 11:28 AM

hey guys i was watchin this movie the other day "21", in this a professor asks his student a question,it goes like this

In a game show we have 3 doors A,B,C behind one of the doors is a CAR and behind other 2 Ghost.
now the host asks u to select one of the door(remember u dont know what is behind what door) and say u select door A, now host tries to play a trick, as he know what is behind each door, he opens door C, and obviously there is a ghost behind the door, now if he asks u if u wish to change ur selection i.e . do u wish to go for B ?????????
what should u say to maximise ur chance of winning, should u go for B or should u stick to A???


in the movie they have told what u should do to maximise ur chances but they have not explained y.
think over it puys i know the answer as i have seen the movie, but the best part is i know y it is so, i was thinking over it yesterday, so thought of sharing it with u guys

pour in ur answers, lets c if u can get it right!!!!
dont just post the answer, if possible post the explanation also and the probability for him to win


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Last edited by Green_Boy; 07-08-2008 at 12:39 PM.
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Re: Probability question - 07-08-2008, 02:18 PM

Quote:
Originally Posted by Green_Boy View Post
hey guys i was watchin this movie the other day "21", in this a professor asks his student a question,it goes like this

In a game show we have 3 doors A,B,C behind one of the doors is a CAR and behind other 2 Ghost.
now the host asks u to select one of the door(remember u dont know what is behind what door) and say u select door A, now host tries to play a trick, as he know what is behind each door, he opens door C, and obviously there is a ghost behind the door, now if he asks u if u wish to change ur selection i.e . do u wish to go for B ?????????
what should u say to maximise ur chance of winning, should u go for B or should u stick to A???

in the movie they have told what u should do to maximise ur chances but they have not explained y.
think over it puys i know the answer as i have seen the movie, but the best part is i know y it is so, i was thinking over it yesterday, so thought of sharing it with u guys

pour in ur answers, lets c if u can get it right!!!!
dont just post the answer, if possible post the explanation also and the probability for him to win
He has said in the movie that the probability of winning increases if the player taked the switch. Can be better explained like this.
There are 3 doors and there is a car behind one and 2 goats behind the other 2 doors. There can 3 kinds of plcaings 1)CGG 2)GCG 3) GGC
Lets consider each case assuming the player always opts for door1. Same can extended to all doors.
1) Player chooses 1. Host opens door 3
If swithced- loses
no switch - wins
* case remains same even if host opnes door 2
2)Player chooses door 1. host opens door 3
if switched - Wins
not switched - loses

3) player chooses door 1. Host opens door 2
If switched - wins
not switched - loses

In 3 cases he wins 2 if he switches his choice
In 3 cases he loses 2 if not switched.
Probablity increases if switched

( Host knows where the car is and he opens the door where there is a goat)


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Green_Boy Green_Boy is offline
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Re: Probability question - 07-08-2008, 03:20 PM

Quote:
Originally Posted by manu hoysala View Post
He has said in the movie that the probability of winning increases if the player taked the switch. Can be better explained like this.
There are 3 doors and there is a car behind one and 2 goats behind the other 2 doors. There can 3 kinds of plcaings 1)CGG 2)GCG 3) GGC
Lets consider each case assuming the player always opts for door1. Same can extended to all doors.
1) Player chooses 1. Host opens door 3
If swithced- loses
no switch - wins
* case remains same even if host opnes door 2
2)Player chooses door 1. host opens door 3
if switched - Wins
not switched - loses

3) player chooses door 1. Host opens door 2
If switched - wins
not switched - loses

In 3 cases he wins 2 if he switches his choice
In 3 cases he loses 2 if not switched.
Probablity increases if switched

( Host knows where the car is and he opens the door where there is a goat)
thats correct!!!!!!
i also got a msg from a puy mentioning this link Monty Hall problem - Wikipedia, the free encyclopedia.
i didnt know it was such a famous problem , but that i arrived at the solution all by myself makes me feel gud
cheers


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vickythakre vickythakre is offline
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Re: Probability question - 04-09-2008, 04:54 PM

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Originally Posted by supervish View Post
yayayayayayayayaya !! my answer's right !



112/64C2 = 1/18

Me dont have 100% track record with probability, so this helps

Vish

hi...how the sample space will be 64C2 when we can arrange 3 squares in 3 form so 27 such squares can be find out
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Re: Probability question - 20-09-2008, 05:13 PM

Megha Fabricators purchased a used milling machine from Keshav Engineering Works. It wascommunicated to Megha Fabricators that there would be a probability of 5% that the outputfrom the machine will bre defective. It is expected that this rate of defective output will continue in future. What will be the probability that maximum one out of every ten jobs done in thatmachine will be defective?


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Re: Probability question - 20-09-2008, 06:26 PM

Quote:
Originally Posted by aviva View Post
Megha Fabricators purchased a used milling machine from Keshav Engineering Works. It wascommunicated to Megha Fabricators that there would be a probability of 5% that the outputfrom the machine will bre defective. It is expected that this rate of defective output will continue in future. What will be the probability that maximum one out of every ten jobs done in thatmachine will be defective?

The probability of none of the jobs to be defective.i.e.,all the jobs are perfect depends on the given probability,i.e.,95% and is

(19/20)(19/20)(19/20)...10 times= 19 to power of 10/20 to power of 10
Similarly,the probability of exactly one job is defective is
19/20.....9 times x 10 x1/20
The fimal value is 19 to power of 9 x 29/20 to power of 10
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Re: Probability question - 22-10-2008, 12:42 PM

A and B throw one dice for a stake of Rs 11 which is to be won by the player who first throws a six.The game ends when the stake is won by A or B.If A has the first thow , what are their respective expectations.
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Re: Probability question - 31-10-2008, 03:10 PM

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Originally Posted by chinturocks View Post
A and B throw one dice for a stake of Rs 11 which is to be won by the player who first throws a six.The game ends when the stake is won by A or B.If A has the first thow , what are their respective expectations.
prob of wining of A = 1/6 + (5/6)^2.(1/6) + (5/6)^4.(1/6) + .........infinity
= 1/6 [ 1/(1-25/36)]
= 1/6* 36/11 = 6/11
prob of wining of B = 1-6/11 = 5/11

so A can expect to win Rs 6/11*11 = 6 and B = 5


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Probability question - 31-10-2008, 04:18 PM

Question:-Two friends decide to meet at a place between 4-5 o clock..
anyone who comes there first waits for the other for 10 mins and then leaves..
whats the probability that they will meet??

Please solve it and tell me the approach.
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