last minute help please!

[Angelz369]

I've done these type of questions before but I can't seem to find the value of X!!
The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is
A) 8
B) 9
C) 10
D)11
E) 12
thankyou!!!

[geoff_821]

Probability of head (lol) = 0.5

therefore  0.5^n = 0.0005
loge(0.5)^n = loge(0.0005)
n = loge(0.0005)/loge(0.5)

you could also use log10 instead of loge

[Angelz369]

Probability of head (lol) = 0.5

therefore  0.5^n = 0.0005
loge(0.5)^n = loge(0.0005)
n = loge(0.0005)/loge(0.5)

you could also use log10 instead of loge

thx!
But if I want to use Binomial on the calculator what values do I put in?
Like I know the first two is BinomPdf (x,0.5,?)-what do I put in for the ? spot
I tried 0.5 but it doesn't work

[panicatthelunchbar]

On CAS: ncr(n,n)(0.5)^n (0.5)^0 which gives you (0.5)^n and you solve this for <0.0005 and it should give you something like 10.96... which is 11

[Angelz369]

On CAS: ncr(n,n)(0.5)^n (0.5)^0 which gives you (0.5)^n and you solve this for <0.0005 and it should give you something like 10.96... which is 11

Yes!! I finally got it!
Thankyou!!!