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
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
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
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@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
@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,
@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
@Dave: Thanks for the explanation :)
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For m
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
@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 *
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 *
@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 *
@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/
@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
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
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
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
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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.
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@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
plz reply am i right or wrong
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For more opt
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)
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@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
@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
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@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
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no then it will be 1/2
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For more options, v
Just to resolve the issue what will be the probability of getting 6
consecutive heads
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@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
@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
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
@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
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
@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
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
@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
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.
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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
@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
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
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
@don: i too get yr answer 17/18 using conditional probability...does that
make sense??i guess this is first new answer lol
On Mon, Aug 8, 2011 at 9:29 PM, Don wrote:
> The answer is 17 in 18, because flipping 5 heads in a row is evidence
> that the probability is high that we have the coin with
The answer is 17 in 18, because flipping 5 heads in a row is evidence
that the probability is high that we have the coin with two heads.
Don
On Aug 7, 12:34 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 t
I think the answer is 17/80, because
as you say the 5 trials are independent.. but
the fact that a head turns up in all the 5 trials, give some
information about our original probability of choosing the coins.
in case we had obtained a tail in the first trial, we can be sure its
the fair coin, and
I think 17/80 is wrong
because if you say that while calculating the answer 3/5, you havent
included the first 5 cases, then even after including it will only increase
the probability of getting the biased coin in hand and thus increasing the
overall probability of getting the heads and 17/80 is a
@brijesh
*first five times* is mentioned intentionally to mislead i think. I vote for
3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong.
On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur wrote:
> (3/5)
>
> On Aug 7, 10:34 pm, Algo Lover wrote:
> > A bag contains 5 coins. Fo
(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 coin from the bag and tossed it 5
> times, heads turned up all five times. What is the probability that
> you toss next time, heads turns up. (All
I think the answer is 3/5, becoz all the trails/tossing coins are
independent events. So even when it is 100th time the answer is 3/5.
On 8 August 2011 09:05, Kamakshii Aggarwal wrote:
> i think the answer will be 1/5+ 4/5*1/2=3/5
>
> coz the question is saying what is the probability of gettin
i think the answer will be 1/5+ 4/5*1/2=3/5
coz the question is saying what is the probability of getting head* sixth
time*(when 5 heads were already there)..
it is not saying what is the probability of getting heads* 6 times*..(in
this case answer will be 17/180)
On Mon, Aug 8, 2011 at 9:54 A
1/5+1/2^6
On Aug 7, 8:34 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 time, heads turns up. (
I think 17/80 is right answer.. otherwise no use of mentioning *first five
times* specifically in the question. ! though m not sure
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@Puneet: So are you saying that 100 heads in a row wouldn't convince
you that you had the unfair coin? How many heads in a row would it
take?
Dave
On Aug 7, 2:40 pm, Puneet Gautam wrote:
> Sixth toss is independent of previous tosses and dependent only on
> coin selection...!
>
> 1/5 + 4/5(1/2)=
abe yaar kya farak padta hai... 3/5=0.6 , other one may be 0.4 ya 0.3,
0.3 ke difference ke liye lad rahe ho...
Chill guys...
On 8/8/11, Shuaib Khan wrote:
> On Mon, Aug 8, 2011 at 12:51 AM, aseem garg wrote:
>
>> @Shuaib: **What is the probability that you toss *next time, heads turns
>> up
On Mon, Aug 8, 2011 at 12:51 AM, aseem garg wrote:
> @Shuaib: **What is the probability that you toss *next time, heads turns
> up***.
Well if you interpret it your way, then you are right. Otherwise, not.
>
> Aseem
>
>
>
> On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan wrote:
>
>>
>>
>> On Mon,
no fight.. lets mention both the answers :D
On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan wrote:
>
>
> On Mon, Aug 8, 2011 at 12:47 AM, aseem garg wrote:
>
>> Think it like this. I have tossed a coin 5 times and it showed heads all
>> the times. What is the probabilty of it shoing a HEADS now?
>>
@Shuaib: **What is the probability that you toss *next time, heads turns up
***.
Aseem
On Mon, Aug 8, 2011 at 1:19 AM, Shuaib Khan wrote:
>
>
> On Mon, Aug 8, 2011 at 12:47 AM, aseem garg wrote:
>
>> Think it like this. I have tossed a coin 5 times and it showed heads all
>> the times. What
On Mon, Aug 8, 2011 at 12:47 AM, aseem garg wrote:
> Think it like this. I have tossed a coin 5 times and it showed heads all
> the times. What is the probabilty of it shoing a HEADS now?
> Aseem
>
Well you are thinking about it the wrong way. Question asks that what is the
probability that head
Think it like this. I have tossed a coin 5 times and it showed heads all the
times. What is the probabilty of it shoing a HEADS now?
Aseem
On Mon, Aug 8, 2011 at 1:12 AM, Shuaib Khan wrote:
>
>
> On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam wrote:
>
>> Sixth toss is independent of previous t
On Mon, Aug 8, 2011 at 12:40 AM, Puneet Gautam wrote:
> Sixth toss is independent of previous tosses and dependent only on
> coin selection...!
>
> 1/5 + 4/5(1/2)= 3/5
>
> is the correct answer
>
> we want to calc. probability of getting heads the sixth time only
> even if it would have be
Sixth toss is independent of previous tosses and dependent only on
coin selection...!
1/5 + 4/5(1/2)= 3/5
is the correct answer
we want to calc. probability of getting heads the sixth time only
even if it would have been 100 th time...3/5 would be the answer
only..
On 8/8/11, Prakash D
1.) coin is fair
2.) coin is unfair
P(head) for unfair coin= 1/5 * 1= 1/5
P(head) for fair coin= 4/5* 1/2 = 2/5
the probability at any instant that the tossed coin is a head is 3/5
17/80 is the probability to get head at all the six times.
the soln. for this problem will be 3/5
On Mon, Aug 8,
If the coin is unbiased then probability of heads: 1/2 irrespective of
whether it is first time or nth time. So answer should be 3/5.
Aseem
On Mon, Aug 8, 2011 at 12:39 AM, saurabh chhabra wrote:
> Even u dont get why u people are gettin 17/80...the probability that
> it will be a head 6th time
Even u dont get why u people are gettin 17/80...the probability that
it will be a head 6th time will be same as the frst time...so it shud
be 3/5...
On Aug 7, 11:05 pm, Kunal Yadav wrote:
> @algo: We can get head in two cases:-
>
> 1.) coin is biases
> 2.) coin is not biased
>
> P(head) for biase
@algo: We can get head in two cases:-
1.) coin is biases
2.) coin is not biased
P(head) for biased= 1/5 *1*1*1*1*1*1= 1/5
P(head) for unbiased= 4/5*(1/2)^6
hence combined probability is what nitish has already mentioned. Hope you
get the point.
On Sun, Aug 7, 2011 at 11:29 PM, Algo Lover wrote:
17/80..what is the correct answer
On Sun, Aug 7, 2011 at 11:25 PM, saurabh chhabra wrote:
> sry...its wrong
>
> 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
>
Can anyone explain the approach how to solve this .
I think all tosses are independent so it should be 3/5. why is this in-
correct
On Aug 7, 10:55 pm, saurabh chhabra wrote:
> sry...its wrong
>
> On Aug 7, 10:34 pm, Algo Lover wrote:
>
>
>
>
>
>
>
> > A bag contains 5 coins. Four of them are f
sry...its wrong
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 time, heads turn
0.6?
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 time, heads turns up. (All
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