Binomial Help (1 Viewer)

[popjin]

Hi,

I'm having some trouble with C) iii, I'm not sure how to find the number of weeks outside of guessing and checking.

Here's what I've evaluated so far:

P(X=2) = nC2 * (9/10)^n-2 * 1/100
P(X=3) = nC3 * (9/10)^n-3 * 1/1000

 

[Drdusk]

Hi,

I'm having some trouble with C) iii, I'm not sure how to find the number of weeks outside of guessing and checking.

Here's what I've evaluated so far:

P(X=2) = nC2 * (9/10)^n-2 * 1/100
P(X=3) = nC3 * (9/10)^n-3 * 1/1000

View attachment 28250

Assuming your expressions are correct.

All you need to do is solve for n in the equation P(x=3) > P(x=2)

 

[popjin]

Assuming your expressions are correct.

All you need to do is solve for n in the equation P(x=3) > P(x=2)

I've simplified it to nC2 * 9^n-2 / 10^n-2 < nC3 * 9^n-3 / 10^n-3 * 1/10 but I'm not sure how to proceed further

 

[fan96]

I've simplified it to nC2 * 9^n-2 / 10^n-2 < nC3 * 9^n-3 / 10^n-3 * 1/10 but I'm not sure how to proceed further

Divide by :



Now use



so



Try the rest for yourself. There are a lot of 's you can cancel out there.