@dave - method is right, calculation mistake on my part, getting 17/18
only. thanks anyways.

On Aug 9, 6:26 pm, Dave <[email protected]> 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 * (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 (2/5) / (3/5), which is
> 2/3.
>
> If you didn't know the outcome of the first roll, the probability of
> heads on the second roll would still be 3/5.
>
> Dave
>
> On Aug 9, 2:57 am, Arun Vishwanathan <[email protected]> wrote:
>
>
>
>
>
>
>
> > @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 first toss and head
> > in second toss with unfair coin)? is it due to the fact that we are not
> > looking at the same sample space in both cases?i am not able to visualise
> > the difference in general..this is also the reason why most of the people
> > said earlier 17/80 as the answer....
>
> > moreover, if the question was exactly the same except in that it was NOT
> > mentioned that heads occurred previously , what would the prob of getting a
> > head in the second toss?
>
> > would it be P( of getting tail in first toss and head in second toss given
> > that fair coin is chosen) +P( of getting head in first toss and head in
> > second toss given that fair coin is chosen) +P( getting heads in first toss
> > and heads in second toss given that unfair coin is chosen) ? this for any
> > toss turns out to be 3/5 ....can u explain the logic abt why it always gives
> > 3/5?
>
> > On Tue, Aug 9, 2011 at 7:37 AM, raj kumar <[email protected]> wrote:
> > > plz reply am i right or wrong
>
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