Re: [algogeeks] Re: probability

@Sukun: these probabilities should have been given in the questn.. since it's not given and there are 3 possible answers, i considered probability of each one of them to b correct as 1/3 (since there are 3 possible answers and only one answer is correct) On Sun, Sep 9, 2012 at 12:48 AM, Sukun Tara

Re: [algogeeks] Re: probability

How do you get P(0.25 being correct) = P(0.5 being correct) = P(0.6 being correct) = 1/3? On Saturday, September 8, 2012 4:42:51 PM UTC+5:30, Shruti wrote: > > shouldn't it b done like this : > > P(correct answer when choosing randomly )= P(choosing 0.25)*P(0.25 being > correct ans)+ P(choosing

[algogeeks] Re: probability

It cannot be either 0.25 or 0.50 http://math.stackexchange.com/questions/76491/multiple-choice-question-about-the-probability-of-a-random-answer-to-itself-bein On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: > > [image: -]14Answers >

Re: [algogeeks] Re: probability

shouldn't it b done like this : P(correct answer when choosing randomly )= P(choosing 0.25)*P(0.25 being correct ans)+ P(choosing 0.60)*P(0.60 being correct ans) + P(choosing 0.50)*P(0.50 being correct ans) = 2/4*1/3 + 1/4*1/3 + 1/4*1/3 [since 3 possible answers, hence prob of an ans bei

[algogeeks] Re: probability

Ans : 0.5 there is two cases : i) if correct ans is 0.25 probability will be 2/4 = 0.5 ii) if correct ans is 0.6 or 0.50 probability will be 1/3 = 0.33 since 0.33 is not in option so correct answer will be 0.5. On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: > > [image:

[algogeeks] Re: probability of winning with two cards

@Don, I think i misunderstood the question again.. :).. One of major things that i went wrong with was that for me the deck consisted of 13*3=39 cards..( basically an assumption made based on the way i understood the question) Thanks for the explanation.. On Jan 23, 9:17 pm, Don wrote: > Close.

[algogeeks] Re: probability of winning with two cards

Close. You actually have to be sure that all 6 cards dealt to the players are unique. For instance, if I get 3 points, if you don't require that all the cards dealth in the game are unique, you would conclude that there is a very small, but positive probability that I will win. In reality, 3 points

[algogeeks] Re: probability of winning with two cards

@Don and Sundi.. As Don pointed out, all we are looking for is: sum of a1 > sum of a2 sum of a1 > sum of a3 Assumption: 1) The 2 cards picked for a particular player are unique. 2) Cards are numbered : 1,..., 12, 13. Hence, the following code should give the answer for the a1's probability to

[algogeeks] Re: probability of winning with two cards

@Don.. Yup, it seems I misread it ... :) .. Thanks On Jan 23, 9:17 am, Don wrote: > I think that you are misreading the problem. A1 wins if his sum is > larger than A2's sum and larger than A3's sum. A1's sum doesn't have > to be larger than A2+A3. > Don > > On Jan 22, 5:18 pm, Lucifer wrote: >

[algogeeks] Re: probability of winning with two cards

I think that you are misreading the problem. A1 wins if his sum is larger than A2's sum and larger than A3's sum. A1's sum doesn't have to be larger than A2+A3. Don On Jan 22, 5:18 pm, Lucifer wrote: > @sundi.. > > Lets put is this way.. > > Probability of (a1 wins + a1 draws + a1 losses) = 1, >

[algogeeks] Re: probability of winning with two cards

@sundi.. Lets put is this way.. Probability of (a1 wins + a1 draws + a1 losses) = 1, Now, sample count a1 wins = 46298 ( using the above given code) Hence, the probability (win) = 46298/474552 = .097561 [ @ Don - as i mentioned in my previous post that i had initially missed a f

[algogeeks] Re: probability of winning with two cards

Hi Lucifer, Have you checked the sum of probability of (a winning + b winning + c winning + draw)==1 ? On Jan 22, 2:38 pm, Lucifer wrote: > @above > > editing mistake.. (btw the working code covers it) > /* > int j =*1*; > for(int i = 0; i < 12 ; i+=2) > { >     A[i] = A[i+1] = A[22-i]

[algogeeks] Re: probability of winning with two cards

@above editing mistake.. (btw the working code covers it) /* int j =*1*; for(int i = 0; i < 12 ; i+=2) { A[i] = A[i+1] = A[22-i] = A[21-i] = j; ++j; } */ On Jan 22, 6:53 pm, Lucifer wrote: > @Don.. > > Well i will explain the approach that i took to arrive at the > probability.. > Well ye

[algogeeks] Re: probability of winning with two cards

@Don.. Well i will explain the approach that i took to arrive at the probability.. Well yes u are correct in saying that it doesn't make a lot of sense but then the no. of wins by a1 keeping in mind that a1 > a2 + a3 is much less than a1 <= a2 + a3.. Or may be I have gone wrong in calculating the

[algogeeks] Re: probability of winning with two cards

You are saying that a1 wins roughly 1 in 20 times? How does that make any sence? Don On Jan 19, 2:35 pm, Lucifer wrote: > @correction: > > Probalilty (a1 wins) = 24575/474552 = .051786 > > On Jan 20, 1:30 am, Lucifer wrote: > > > > > hoping that the cards are numbered 1,2,3,,13.. > > > Proba

[algogeeks] Re: probability of winning with two cards

@correction: Probalilty (a1 wins) = 24575/474552 = .051786 On Jan 20, 1:30 am, Lucifer wrote: > hoping that the cards are numbered 1,2,3,,13.. > > Probalilty (a1 wins) = 21723/474552 = .045776 > > On Jan 20, 12:47 am, Don wrote: > > > > > > > > > P= 8800/28561 ~= 0.308112461... > > > On Ja

[algogeeks] Re: probability of winning with two cards

hoping that the cards are numbered 1,2,3,,13.. Probalilty (a1 wins) = 21723/474552 = .045776 On Jan 20, 12:47 am, Don wrote: > P= 8800/28561 ~= 0.308112461... > > On Jan 18, 7:40 pm, Sundi wrote: > > > > > > > > > there are 52 cards.. there are 3 players a1,a2,a3 each player is given > > 2

[algogeeks] Re: probability of winning with two cards

P= 8800/28561 ~= 0.308112461... On Jan 18, 7:40 pm, Sundi wrote: > there are 52 cards.. there are 3 players a1,a2,a3 each player is given > 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of > cards is greater then the other two players sum. > > find the probability of a1 being t

[algogeeks] Re: probability of winning with two cards

@Sunny: The probability of a1 being the winner is not 1/3 because of ties. I.e., if a1 = a2 > a3, then a1 and a2 are tied and there is no winner. What we can say with no calculations is that P(a1 winning) = (1 - P(no winner)) / 3. Dave On Jan 18, 10:52 pm, sunny agrawal wrote: > isn't the answer

[algogeeks] Re: probability of winning with two cards

:)... Lets say you have two players a and b one card is distrbuted to each player if the card with a is higher then a wins else a loses. probability of a winning: total num of cards with combos where a wins a is the first player: 2-1,3-1...13-1 3-2...13-2 .. 13-12 this sum equals = (13+12+...

Re: [algogeeks] Re: probability question

Take it simple silly ... for each 10 min interval, if man comes in first 2 min, he'll catch the 1st train, if he comes in next 8 min, he'll catch the 2nd train. hence for harbor line - (2/10) 0.2 and for main line 0.8. On Wed, Aug 31, 2011 at 10:37 PM, ravi wrote: > > A man goes to the statio

[algogeeks] Re: probability ques

@Aditya: If you are comfortable with working with fractions of area, with area being a real number, then why not length, also being a real number? The probability of a randomly selected point in the interval [1, 100] satisfying a + 100/a < 50 is exactly sqrt(2100)/99. Dave On Sep 1, 3:43 am, Adit

Re: [algogeeks] Re: probability ques

the space in tht case is restricted... my point is...number of points in a 4 m2 area wud be exactly 4 times of the number of points in 1 m2 area... so we r actually talking over area in ur qn...now if i were to say...tht find the probability tht it will hit the target at coordinate (0,0) ... tht wu

Re: [algogeeks] Re: probability question

A man goes to the station every day to catch the first train that comes ?? => man catches the first train that comes after him On Wed, Aug 31, 2011 at 10:30 PM, Dave wrote: > @Ankul: According to the problem statement, the first train is the one > that arrives at 5:00 a.m. > > Dave > > On Aug 3

[algogeeks] Re: probability question

@Ankul: According to the problem statement, the first train is the one that arrives at 5:00 a.m. Dave On Aug 31, 11:26 pm, Ankuj Gupta wrote: > I could not get it. What does first train mean here? > > On Sep 1, 1:08 am, Don wrote: > > > > > Assuming that the man arrives at a random time during

[algogeeks] Re: probability question

I could not get it. What does first train mean here? On Sep 1, 1:08 am, Don wrote: > Assuming that the man arrives at a random time during the 24-hour day, > there are 228 minutes in the day when the next train will be the > harbour line (2 minutes of every 10 for 19 hours). For the other 1212 >

[algogeeks] Re: probability ques

@Aditya. There are an infinite number of points in that target, too. But we don't have any trouble saying that 1/4 of them are in the bullseye. Dave On Aug 31, 4:07 pm, Aditya Virmani wrote: > @DAVE again u r considering a finite space... in the above case...but how > wud u take the space in rea

Re: [algogeeks] Re: probability ques

@DAVE again u r considering a finite space... in the above case...but how wud u take the space in real number thing...with no particular info thr r infinite real number frm 1-100...if i cud change the qn to find the probability the chosen number is in the range a to a+1 ...thn it cud be aptly

[algogeeks] Re: probability question

Assuming that the man arrives at a random time during the 24-hour day, there are 228 minutes in the day when the next train will be the harbour line (2 minutes of every 10 for 19 hours). For the other 1212 minutes the main line will be the next train. Therefore, the probability of catching the main

Re: [algogeeks] Re: probability question

i think ur question is not clear dear acc to ur question the man should reacch the station exactly at 5 or before 5 to get the first train . then prob will be 1 otherwise prob will be 0. correct me if i am wrong. -- You received this message because you are subscribed to the Google

[algogeeks] Re: probability question

@Swetha: My probability of reaching any train station by 5:00 a.m is zero. So I would never catch the first train. Now if the trains started at 8 or 9 a.m., then I might have some probability of catching the first one. :-) (In case it isn't obvious, I'm focusing on the wording of the question at t

[algogeeks] Re: probability question

In my experience, the probability that a train stays on schedule to within 5 minutes is about 0.01, so I'm going to say that the probability is about 0.51. Don On Aug 31, 8:37 am, swetha rahul wrote: > In a railway station, there are two trains going. One in the harbour line > and one in the main

[algogeeks] Re: probability question

0.8 On Aug 31, 6:37 pm, swetha rahul wrote: > In a railway station, there are two trains going. One in the harbour line > and one in the main line, each having a frequency of 10 minutes. The main > line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A > man goes to the statio

Re: [algogeeks] Re: probability ques

Hi Guys, Comment required from NIT Warangal ppls. Who is this guy?? Who claims this ... http://timesofindia.indiatimes.com/tech/careers/job-trends/Facebook-hires-NIT-Warangal-student-for-Rs-45-lakh/articleshow/9793300.cms -- You received this message because you are subscribed to the Google

[algogeeks] Re: probability ques

@AnikKumar: Most people normally wouldn't have difficulty with probabilities on the real numbers. E.g., there is a target with two regions, the bullseye with radius 1 and a concentric region with radius 2. What is the probability of a randomly-thrown dart hitting the bullseye, given that it hits th

Re: [algogeeks] Re: probability ques

Agree with Don. But what if we want to find probability of on real line? How we can consider R as sample space? Is that Sample space should be COUNTABLE and FINITE? *By the quadratic formula, a is 2.08712 or 47.9128. The range is 45.8256. A falls in the range of 1..100 or 99. So the probability

Re: [algogeeks] Re: probability ques

+1 Don... nthin is specified fr the nature of numbers if thy can be rational or thy hav to be only natural/integral numbers... On Wed, Aug 24, 2011 at 9:33 PM, Don wrote: > First find the endpoints of the region where the condition is met: > > a + 100/a = 50 > a^2 - 50a + 100 = 0 > By the quadra

[algogeeks] Re: probability ques

First find the endpoints of the region where the condition is met: a + 100/a = 50 a^2 - 50a + 100 = 0 By the quadratic formula, a is 2.08712 or 47.9128. The range is 45.8256. A falls in the range of 1..100 or 99. So the probability is 47.9128/99 = 0.48397 Don On Aug 23, 11:56 am, ramya reddy wr

Re: [algogeeks] Re: probability ques

ok thanks On Wed, Aug 24, 2011 at 10:02 AM, Dave wrote: > @Dheeraj: Because there are 98 numbers between 1 and 100. They are 2, > 3, 4, ..., 99. It is a matter of semantics. 1 and 100 are not between > 1 and 100. > > Dave > > On Aug 23, 11:21 pm, Dheeraj Sharma > wrote: > > y its 45/98 and not

[algogeeks] Re: probability ques

@Dheeraj: Because there are 98 numbers between 1 and 100. They are 2, 3, 4, ..., 99. It is a matter of semantics. 1 and 100 are not between 1 and 100. Dave On Aug 23, 11:21 pm, Dheeraj Sharma wrote: > y its 45/98 and not 45/100 ?? > > On Wed, Aug 24, 2011 at 9:41 AM, Manoj Bagari wrote: > > oh

[algogeeks] Re: probability ques

sorry i thought it was a+ 100/a>50 because same question was asked by national instrument 2 days ago in this case it'll be [3-47] 45 no so prob will be 45/100 On Aug 24, 8:21 am, Ankur Goel wrote: > Dude how it can be 50 if u do 50 + 100/50 which is 52> 50 > > > > On Wed, Aug 24, 2011 at 8:42 AM

Re: [algogeeks] Re: probability! tough one to crack!

a wonderful explaination divye!! thank you so much :) :) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@goo

Re: [algogeeks] Re: probability! tough one to crack!

GOOD one dave and thanks divye for a wonderful explanation On Fri, Aug 19, 2011 at 11:44 PM, DK wrote: > For those of you who want an explanation of Dave's answer, please refer to > the diagram below. > > | 0.5 in |-| 0.5 in | > xxx - > xxx

Re: [algogeeks] Re: probability! tough one to crack!

For those of you who want an explanation of Dave's answer, please refer to the diagram below. | 0.5 in |-| 0.5 in | xxx - xxx 0.5 in xxx - xx...xx | xx...xx | x

Re: [algogeeks] Re: probability! tough one to crack!

got it. Sanju :) On Fri, Aug 19, 2011 at 10:25 AM, Romil ... wrote: > Because this is the answer. Rest of the times it will not touch any of the > grid lines. > > > On Fri, Aug 19, 2011 at 10:50 PM, priya ramesh < > love.for.programm...@gmail.com> wrote: > >> why this sentence?? >> >> "3/4

Re: [algogeeks] Re: probability! tough one to crack!

Because this is the answer. Rest of the times it will not touch any of the grid lines. On Fri, Aug 19, 2011 at 10:50 PM, priya ramesh < love.for.programm...@gmail.com> wrote: > why this sentence?? > > "3/4 of > the time the coin will touch a grid line"?? > > -- > You received this message becaus

Re: [algogeeks] Re: probability! tough one to crack!

why this sentence?? "3/4 of the time the coin will touch a grid line"?? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks

Re: [algogeeks] Re: probability! tough one to crack!

The requirements will be satisfied if the centre of the coin is such that the coin just touches the square. This is possible only when the centre of coin is in a smaller square of 1 inch side. Hence the result. On Fri, Aug 19, 2011 at 10:41 PM, Sanjay Rajpal wrote: > @Dave : me too didnt get t

Re: [algogeeks] Re: probability! tough one to crack!

@Dave : me too didnt get the meaning you want to convey, plz throw some light. Sanjay Kumar B.Tech Final Year Department of Computer Engineering National Institute of Technology Kurukshetra Kurukshetra - 136119 Haryana, India On Fri, Aug 19, 2011 at 10:09 AM, priya ramesh < love.for.programm..

Re: [algogeeks] Re: probability! tough one to crack!

@dave: You are great!! The ans is indeed 1/4. I dint understand this sentence... "The area of the region within 1/2 inch of the boundary is 3 square inches." -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email t

[algogeeks] Re: probability! tough one to crack!

@Priya: Consider one of the squares in the grid. It has an area of 4 square inches. If the coin lands so that its center is within 1/2 inch of any grid line, the coin will touch the line. The area of the region within 1/2 inch of the boundary is 3 square inches. Therefore, 3/4 of the time the coin

Re: [algogeeks] Re: probability tough one!

Sorry the problems are not same.I should have read the problem more carefully.Anyways I would recommend its high time you get a hard copy of coreman.. Trying for a solution now. On Wed, Aug 17, 2011 at 7:19 PM, priya ramesh wrote: > I don't have coreman. If you have an e book can you plz uploa

Re: [algogeeks] Re: probability tough one!

I don't have coreman. If you have an e book can you plz upload it?? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsu

Re: [algogeeks] Re: probability tough one!

The question is directly taken from coreman...read it,the best is explained there On Wed, Aug 17, 2011 at 7:10 PM, Adi Srikanth wrote: > > there is something anamoly about this birthday probability caculation. Search > in google..you may find it. > > > Regards, > Adi Srikanth. > Mob No 988723334

Re: [algogeeks] Re: probability tough one!

there is something anamoly about this birthday probability caculation. Search in google..you may find it. Regards, Adi Srikanth. Mob No 9887233349 Personal Pages: adisrikanth.co.nr On Wed, Aug 17, 2011 at 6:25 PM, Romil ... wrote: > Take it as: > P(atleast 2) = 1-P(no 2 have same b'day) =

[algogeeks] Re: probability tough one!

http://en.wikipedia.org/wiki/Birthday_problem On Aug 17, 5:55 pm, "Romil ..." wrote: > Take it as: > P(atleast 2) = 1-P(no 2 have same b'day) = 1- ((365C50)/50!) > where C represents the combinations > > On Wed, Aug 17, 2011 at 6:14 PM, sukran dhawan wrote: > > > > > > > can yo explain it pl?

Re: [algogeeks] Re: probability tough one!

Take it as: P(atleast 2) = 1-P(no 2 have same b'day) = 1- ((365C50)/50!) where C represents the combinations On Wed, Aug 17, 2011 at 6:14 PM, sukran dhawan wrote: > can yo explain it pl? > > > On Wed, Aug 17, 2011 at 6:11 PM, Aditya Jain wrote: > >> Is that exactly 2 or atleast 2? >> >> P(atleas

Re: [algogeeks] Re: probability tough one!

can yo explain it pl? On Wed, Aug 17, 2011 at 6:11 PM, Aditya Jain wrote: > Is that exactly 2 or atleast 2? > > P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) > > > > > > On Aug 17, 5:24 pm, priya ramesh > wrote: > > nothing is specified. I guess it's 365 > > -- > You rece

[algogeeks] Re: probability tough one!

Is that exactly 2 or atleast 2? P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) On Aug 17, 5:24 pm, priya ramesh wrote: > nothing is specified. I guess it's 365 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To pos

Re: [algogeeks] Re: Probability Puzzle

I'm little late but I too got 17/18. On Tue, Aug 16, 2011 at 10:47 PM, Jacob Ridley wrote: > I think there is some ambiguity in the question. > > >>> (All this time you don't know you were tossing a fair coin or not). > 1) Does the above statement mean that the thower don't know whether he > or s

[algogeeks] Re: Probability Puzzle

I think there is some ambiguity in the question. >>> (All this time you don't know you were tossing a fair coin or not). 1) Does the above statement mean that the thower don't know whether he or she threw a fair coin even after throwing? Or is the thrower not informed beforehand that one of them i

[algogeeks] Re: Probability Puzzle

A=p(biased coin/5 heads)=8/9 probability that the coin is biased given 5 heads (bayes theorem) B=p(unbiased coin/5 heads)=1/9 P(6th head)=A*1+B*1/2=17/18 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to

Re: [algogeeks] Re: Probability

sorry that was for equal case : for unequal case 1-(73/648)= 575/648 On Fri, Aug 12, 2011 at 2:28 AM, Prakash D wrote: > total possible outcomes= 6*6*6*6 > > possibility of sum gives 2 in both pair --> 1*1*1*1 =1 > possibility of sum gives 3 in both pair --> 2*1*2*1 =4 > {because the possibiliti

Re: [algogeeks] Re: Probability

total possible outcomes= 6*6*6*6 possibility of sum gives 2 in both pair --> 1*1*1*1 =1 possibility of sum gives 3 in both pair --> 2*1*2*1 =4 {because the possibilities are (1,2)(1,2), (1,2)(2,1), (2,1)(1,2), (2,1)(2,1)} possibility of 4 --> 9 { 2,2 1,3 and 3,1} possibility of 5 --> 16 {1,4 2,3

[algogeeks] Re: Probability

Let f(x)= no. of ways of getting a sum x when pair of dice is thrown. S()=sum . tot=S( f(i) ) 2<=i<=12 So the solution is : S( f(i)*(tot-f(i)) )/(6^4). Is daT fine.. ? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this gr

Re: [algogeeks] Re: Probability Puzzle

@dave : nice explanationsthank you for pointing out :) On Wed, Aug 10, 2011 at 3:39 AM, Prakash D wrote: > @dave: thank you.. nice explanation :) > > > On Wed, Aug 10, 2011 at 3:24 AM, Dave wrote: > >> @Ritu: We are flipping one coin five times. Are you saying that you >> don't learn anythi

Re: [algogeeks] Re: Probability Puzzle

@dave: thank you.. nice explanation :) On Wed, Aug 10, 2011 at 3:24 AM, Dave wrote: > @Ritu: We are flipping one coin five times. Are you saying that you > don't learn anything about the coin by flipping it? Would you learn > something if any one of the five flips turned up tails? After a tails,

[algogeeks] Re: Probability Puzzle

@Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19

Re: [algogeeks] Re: Probability Puzzle

@Dave: Thanks for the explanation :) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For m

[algogeeks] Re: Probability Puzzle

The statement "You randomly pulled one coin from the bag and tossed" tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover wrote: > A bag contains 5 coins. Four of them are fair and one has heads on > both sides. You randomly pulled one coi

[algogeeks] Re: Probability Puzzle

@dave - method is right, calculation mistake on my part, getting 17/18 only. thanks anyways. On Aug 9, 6:26 pm, Dave wrote: > @Arun: The probability of getting a head on the first toss is > 1/5 * 1 + 4/5 * (1/2) ) = 3/5, > while the probability of getting 2 consecutive heads is > 1/5 * 1 + 4/5 *

[algogeeks] Re: Probability Puzzle

The probability of getting n consecutive heads is P(n heads) = 1/5 * 1 + 4/5 * (1/2)^n, Thus, the probability of getting a head on the n+1st roll given that you have gotten heads on all n previous rolls is P(n+1 heads | n heads) = P(n+1) / P(n) = ( 1/5 * 1 + 4/5 * (1/2)^(n+1) ) / ( 1/5 * 1 + 4/5 *

[algogeeks] Re: Probability Puzzle

@dave- calculation mistake on my part - method is right. getting 17/18 only thanks anyways. On Aug 9, 6:26 pm, Dave wrote: > @Arun: The probability of getting a head on the first toss is > 1/5 * 1 + 4/5 * (1/2) ) = 3/5, > while the probability of getting 2 consecutive heads is > 1/5 * 1 + 4/5 *

[algogeeks] Re: Probability Puzzle

@Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/

[algogeeks] Re: Probability Puzzle

@Arpit: No. The probability of getting 6 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^6 ) = 17/80, while the probability of getting 5 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^6 ) = 9/40. Thus, the probability of getting a head on the sixth roll given that you have gotten heads on all five previous

[algogeeks] Re: Probability Puzzle

ans is 16/17 + 1/2*1/17 = 33/34 On Aug 7, 10:34 pm, Algo Lover wrote: > A bag contains 5 coins. Four of them are fair and one has heads on > both sides. You randomly pulled one coin from the bag and tossed it 5 > times, heads turned up all five times. What is the probability that > you toss next

Re: [algogeeks] Re: Probability Puzzle

go through the posts before posting anything :) On Tue, Aug 9, 2011 at 6:29 PM, arpit.gupta wrote: > it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 > > On Aug 7, 10:54 pm, Nitish Garg wrote: > > Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 > > -- > You received this message because y

[algogeeks] Re: Probability Puzzle

it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg wrote: > Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@g

Re: [algogeeks] Re: Probability Puzzle

If we are selecting a new coin after getting five consecutive heads, then the probability of getting a sixth head is same as getting the first head. But we are tossing the same coin again. So, conditional probability has to be used. -- Regards, Vanathi -- You received this message because yo

Re: [algogeeks] Re: Probability Puzzle

@dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in fir

Re: [algogeeks] Re: Probability Puzzle

plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more opt

Re: [algogeeks] Re: Probability Puzzle

http://math.arizona.edu/~jwatkins/f-condition.pdf see this link now ithink the answer should be 65/66 bcoz the probability of selectting double headed coin after n heads =2^n/2^n+1 and fair coin is =1/2^n+1 so for 6th head it should be :2^n/2^n+1*1+((1/2^n+1)*1/2) -- You received this messag

Re: [algogeeks] Re: Probability Puzzle

@Dave: I guess 17/18 is correct. Since we have to *calculate the probability of getting a head in the 6th flip given that first 5 flips are a head*. Can you please explain how you got the values of consequent flips when you said this? *"In fact, the probability is 3/5 for the first flip. After a h

[algogeeks] Re: Probability Puzzle

@Raj. Good. So now answer my last question? Dave On Aug 9, 12:21 am, raj kumar wrote: > no then it will be 1/2 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe

[algogeeks] Re: Probability Puzzle

@Raj: After getting 5 consecutive heads, the probability of getting a 6th head is 17/18. Dave On Aug 9, 12:17 am, raj kumar wrote: > Just to resolve the issue what will be the probability of getting 6 > consecutive heads -- You received this message because you are subscribed to the Google Gro

Re: [algogeeks] Re: Probability Puzzle

no then it will be 1/2 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, v

Re: [algogeeks] Re: Probability Puzzle

Just to resolve the issue what will be the probability of getting 6 consecutive heads -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email

[algogeeks] Re: Probability Puzzle

@Raj. Granted that the first flip has a 3/5 probability of getting a head. But if it produces a tail, would you say that the second flip also has a 3/5 probability of getting a head? Or have you learned something from the tail? If you learn something from a tail, why don't you learn something from

[algogeeks] Re: Probability Puzzle

@Coder: You (and others) are saying that the probability of a head is 3/5 on the first flip, and that it doesn't change after any number of heads are flipped. Notice, however, that if the first flip were tails, you wouldn't say that the probability of getting heads on the next flip is 3/5. You woul

Re: [algogeeks] Re: Probability Puzzle

Pls check the ques 8th This may remove misunderstanding... http://www.folj.com/puzzles/difficult-logic-problems.htm On Tue, Aug 9, 2011 at 10:21 AM, raj kumar wrote: > @all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 > > > when it's already given that 5 heads have turned up

Re: [algogeeks] Re: Probability Puzzle

@all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the an

Re: [algogeeks] Re: Probability Puzzle

it's 3/5 On Tue, Aug 9, 2011 at 8:29 AM, Dave wrote: > @Dipankar: You are correct about the answer to your alternative > question being 17/80, but your answer 3/5 says that you don't think > you have learned anything by the five heads flips. Don has given a > good explanation as to why the answe

[algogeeks] Re: Probability Puzzle

@Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is non

Re: [algogeeks] Re: Probability Puzzle

3/5. As the question doesn't ask anything about the sequence. Had the question been " Find the probability that all 6 are H " then it would have been 17/80. On 9 August 2011 04:07, Dave wrote: > @Vinay: What if you tossed 100 consecutive heads? Would that be enough > to convince you that you ha

[algogeeks] Re: Probability Puzzle

@Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of gettin

[algogeeks] Re: Probability Puzzle

answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeek

Re: [algogeeks] Re: Probability Puzzle

Man, I feel so stupid. Yes, it is a case of conditional probability. We have to calculate the probability of six heads, "given" that 5 heads have occured. So answer is 17/18. On Tue, Aug 9, 2011 at 1:47 AM, Arun Vishwanathan wrote: > @shady: 3/5 can be the answer to such a question: what is prob

Re: [algogeeks] Re: Probability Puzzle

@shady: 3/5 can be the answer to such a question: what is prob of getting head on nth toss if we have 4 coins fair and one biased...then at nth toss u choose 4/5 1/5 prob and then u get 3/5 @shady , don: i did this: P( 6th head | 5 heads occured)= P( 6 heads )/ P( 5 heads) answr u get is 17/18..i

Re: [algogeeks] Re: Probability Puzzle

answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don wrote: > Consider the 5 * 64 possible outcomes for the selection of coin and > six flips, each one happening with equal probability. Of those 320 > possible outcomes, 4*62 are excluded by knowing that

[algogeeks] Re: Probability Puzzle

Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair

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