# Math Help - Show that p= 0.05

1. ## Show that p= 0.05

The variable X has a binomial distribution with n=20 and probability of success p, where p<0.1.
(i) write down an expression, in terms of p, for P(X=2).

(ii) show that p=0.05, correct to 2 decimal places if given that P(X>=1) [x is larger or equals to 1)=0.641541

Im having problems with part (ii)

This is my attempt at it: If its rubbish, just ignore it! x.x

P(X>=1) = 0.641541
P(X>=1) can be written as 1-P(X=0)
Hence,
1-P(X=0) =0.641541

Given that n=20

20C0 x p^0 x (1-p)^20 = 0.641541
1 x (1-p)^20 = 0.641541
p^20= 0.641541
p= 0.9780483099

Thank you really really much!

2. ## Re: Show that p= 0.05

Originally Posted by pumbaa213
The variable X has a binomial distribution with n=20 and probability of success p, where p<0.1.
(i) write down an expression, in terms of p, for P(X=2).

(ii) show that p=0.05, correct to 2 decimal places if given that P(X>=1) [x is larger or equals to 1)=0.641541

Im having problems with part (ii)

This is my attempt at it: If its rubbish, just ignore it! x.x

P(X>=1) = 0.641541
P(X>=1) can be written as 1-P(X=0)
Hence,
1-P(X=0) =0.641541

Given that n=20

20C0 x p^0 x (1-p)^20 = 0.641541
1 x (1-p)^20 = 0.641541
p^20=0.641541
you were fine up until here.
you want to solve

$(1-p)^{20}=0.641541$

3. ## Re: Show that p= 0.05

I can't seem to solve it. could you help me a little more? thank you so much!

4. ## Re: Show that p= 0.05

really?

$(1-p)^{20}=0.641541$

$1-p=\sqrt[20]{0.641541}$

$p=1-\sqrt[20]{0.641541}$

$p=0.022$

5. ## Re: Show that p= 0.05

yep! i got 0.022 as well but aren't I supposed to show that p=0.05?
But thanks so much for the help, i really appreciate it!

6. ## Re: Show that p= 0.05

Originally Posted by pumbaa213
yep! i got 0.022 as well but aren't I supposed to show that p=0.05?
But thanks so much for the help, i really appreciate it!
I see what the problem is.
We should be solving

$(1-p)^{20}=(1-0.641541)=0.358459$

$p=1-\sqrt[20]{0.358459}=0.05$

7. ## Re: Show that p= 0.05

thank you so so so much romsek!