probablility qu. (1 Viewer)

[currysauce]

How many times must a coin be tossed until the probablility of getting 2 or more heads exceeds 0.99?

cheers

 

[darkliight]

I get 11.

 

[currysauce]

could u elaborate?

 

[currysauce]

10 is the correct answer, as the answer sheet alludes

 

[darkliight]

Well I had 10, just by looking at a pattern, then I thought if that is the right answer then ...

I have 2^10 = 1024 possible combinations. Only 11 of these don't have more than one head, ie.

TTTTTTTTTT
HTTTTTTTTT
THTTTTTTTT
...
TTTTTTTTTH

So the odds of me not getting that (which is what we want) is 1013/1024 ~ 0.98925. Sure that rounds up, but to be safe I just changed my answer to 11. Maybe my reasoning is dodgy though. Sorry that isn't much help.

 

[Nodice]

Hows this:

(n) 0.5^n + (n) 0.5^n = 0.01
(1)...............(0)

But since (n) = n and (n) = 1
...............(1)...............(0)

(0.5^n)[n+1] = 0.01
And by inspection we find that the value is somewhere between 10 and 11.

 

[jarrypan]

Hows this:

(n) 0.5^n + (n) 0.5^n = 0.01
(1)...............(0)

But since (n) = n and (n) = 1
...............(1)...............(0)

(0.5^n)[n+1] = 0.01
And by inspection we find that the value is somewhere between 10 and 11.

I couldn't understand that...

 

[Nodice]

(n) is short hand for nC1.. it uses the binomial expansion theorem for probability.
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