@rajesgeeks:i think probability of sucess of 2nd game is 2(x^2(1-x))+x^2
becoz as explained below:
1.HH.......so success in 2 shots no need of 3rd one
2.HFH
3.FHH
where H-hit the hoop.
F-failed to hit the hoop.
On Wed, Mar 2, 2011 at 7:20 PM, [email protected] <[email protected]
> wrote:
> let p be the probability
>
> second game success probability is
> 3*[x^2(1-x)]+x^2
>
> first game success probability is
> x
>
> find for which x ,3*[x^2(1-x)]+x^2<x,this case game1 preffreed
> and for which x ,3*[x^2(1-x)]+x^2>x,this case game2 preffreed
>
>
>
>
> On Mar 2, 4:15 am, bittu <[email protected]> wrote:
> > You have a basketball hoop and someone says that you can play 1 of 2
> > games.
> > Game #1: You get one shot to make the hoop.
> > Game #2: You get three shots and you have to make 2 of 3 shots.
> > If p is the probability of making a particular shot, for which values
> > of p should you pick one game or the other?
> >
> > Thanks & Regards
> > Shashank
>
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