Binomial distribution please help!

The probabaility that Annie hits the target with a shot is 0.28. Find the least number of shots Annie must fire so that the probabaility of at least one successful shot is greater than 0.99.

Thanks in advance

Reply 1

mastergo

gosh i know this topic sounds boring but please help somebody!

Reply 2

Tla

The first thing you should do is write it out properly. X = number of hits

X ~ B(n, 0.28)
P(x>1) = P(X>=2) = 1 - P(X<=1)

1 - P(X<=1) > 0.99
P(X<=1) < 0.01

Look at your binomial tables for the lowest value of n that satisfies that inequality when p=0.28

Reply 3

mastergo

theres no value of ps that are 0.28... only like 0.25.... 0.300...

Reply 4

Green Clover

mastergo

X~B(n,0.28)
X: the no of successful shots

P(X \geq 1)>0.99

(as the prob of getting at least 1 successful shot is greater than 0.99)

1-P(X=0)>0.99

P(X=0)<0.01

(1-0.28)^n<0.01

n \times ln0.72<ln0.01

n>?

<--- Can you find n now? :smile:

And no, this topic is not boring :p:

Reply 5

mastergo

Green Clover

X~B(n,0.28)
X: the no of successful shots

P(X \geq 1)>0.99

(as the prob of getting at least 1 successful shot is greater than 0.99)

1-P(X=0)>0.99

P(X=0)<0.01

(1-0.28)^n<0.01

n \times ln0.72<ln0.01

n>?

<--- Can you find n now? :smile:

And no, this topic is not boring :p:

oh god, after
P(X=0) < 0.01, I have no idea whats happening, could you explain it please?

Thanks very much

Reply 6

Green Clover

OK so what is the formula you use for finding P(X=r) for the binomial distribution? (without using the table of course :p:

) When r=0, what should the formula be? :smile:

Answer me first then we'll go ahead :wink:

Reply 7

mastergo

nCr x q^n-r x p^r i think it is. when r=0 , nCo x q^n

Reply 8

Green Clover

Yeah so do you understand the first line below P(X=0)<0.01 now? :smile:

From there:

0.72^n<0.01

(q=0.72)
To find n, we use log/ln formula (you can use either log or ln, that won't change the answer). Using ln on both sides gives:

ln0.72^n<ln0.01

n can be taken down so:

n\times ln0.72<ln0.01

Take ln 0.72 to the other side. as 0.72<1, ln0.72<0 (remember for x<1, lnx<0) so we have to change the sign:

n>\frac{ln0.01}{ln0.72}

So can you find n now?
If there's any part you don't understand, just ask :smile:

Reply 9

mastergo

oh ic ok i understand the line below that now: 0.72^n <0.01
and from there... we havent done logs and stuff yet so i wont be able to do it this way :frown:

is there any other way to find the value of n from there?

thanks

Reply 10

Green Clover

Uhm sorry that's how I often solve it. Without using log I don't think it is possible to solve this question when p or at least 1-p is not given in the table(someone please correct me if I'm wrong :o: