Using logarithms in a geometric distribution question involving inequalities.

Dec 2015
25
1
Worcester, UK
The question goes like this:

It is know that 9% of the population belongs to blood group B.

How many people must a doctor - who is sampling random people - examine to be 99.8% confident of finding at least one person with blood group B?

\(\displaystyle P(X\geqx)\leq0.002\)
where X = the no. of people examined until someone with blood group B is found, therefore X has the distribution \(\displaystyle geom(\frac{91}{100}\))

\(\displaystyle \Rightarrow\log_{0.91}0.91^{(x-1)}\leq\log_{0.91}0.0002\)

\(\displaystyle \Rightarrow\(x-1)\leq65.9\) (using the log power rule on the LHS and 'de-logging' the RHS)

\(\displaystyle x\leq66.9\)

However, this should be \(\displaystyle x\geq66.9\) if the above statements are to hold true (and the question asks 'at least,' therefore a 'greater than or equal sign' must be in the final statement.)

Where have I gone wrong; I haven't divided or multiplied by negative numbers, unless unintentionally?

Thanks. :)
 
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Jun 2008
1,389
513
Illinois
Your problem starts here:

\(\displaystyle \Rightarrow\log_{0.91}0.91^{(x-1)}\leq\log_{0.91}0.0002\)
If you start with

\(\displaystyle 0.91^x \leq 0.002 \)

and take log base 0.91 of both sides you have to turn the inequality around because you are converting both sides into negative numbers. It has the same effect with respect to the direction of the inequality as multiplying by -1. So your next step should be:

\(\displaystyle log_{0.91} (0.91^x) \ge \log_{0.91} (0.002) \)

Follow this through and you get \(\displaystyle x \ge 65.9\).

In addition - I don't understand why you have x-1 instead of just plain x in your expression. Also, a typo: you're looking for 1 - 99.8% = 0.002. not 0.0002.
 
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Dec 2015
25
1
Worcester, UK
"Your problem starts here:

If you start with

\(\displaystyle 0.91^x \leq 0.002 \)

and take log base 0.91 of both sides you have to turn the inequality around because you are converting both sides into negative numbers."

Why does taking log base 0.91 convert both sides into negative numbers? \(\displaystyle \log_{0.91}(0.002)\) yields a positive number (namely 65.9). Same with the LHS.
 
Dec 2015
25
1
Worcester, UK
In addition, I believe I used x-1 as an exponent as \(\displaystyle P(X\leq\ x)\Rightarrow\) we have failed to get a non-group B sample a minimum of (x-1) times (however, I have checked the answers in the back and it seems that using x rather x-1 is the correct method for this question (as my answer is one out). What is wrong with my reasoning?
 

ChipB

MHF Helper
Jun 2014
305
124
NJ
Using x rather than x-1 is because the probability that there are no type B people in a sample of x people is 0.91^x. You can see this makes sense by considering the case of x=1: if you have one person there is a 0.91^1 = 91% chance that the person is not type B. Using x-1 would say that in a sample of 1 there is 0.91^0 = 100% chance that the person is not type B - this is clearly not correct.

Now back to the first question - why the inequality has to change. You are correct, log_0.91 (0.002) is a positive value. So the reason lies elsewhere. One thing to notice is that the function 0.91^x has a negative slope, so as x gets bigger 0.91^x gets smaller, and eventually falls below the constant 0.002. But after taking the log of both sides the resulting LHS has an upward slope, whereas log(0.002) is still a constant, so as x gets bigger it eventually exceeds 0.002. The analogy I can think of is if you take the inverse of both sides of an inequality you have to switch the inequality. For example, given x > C if you take the inverse of bith sides you need to switch the inequality to get 1/x < 1/C.
 
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Dec 2015
25
1
Worcester, UK
Using x rather than x-1 is because the probability that there are no type B people in a sample of x people is 0.91^x. You can see this makes sense by considering the case of x=1: if you have one person there is a 0.91^1 = 91% chance that the person is not type B. Using x-1 would say that in a sample of 1 there is 0.91^0 = 100% chance that the person is not type B - this is clearly not correct.
I do apologise; I've realised I've made a mistake when typing this; it is meant to be \(\displaystyle P(X\geq\ x)\) rather than less than or equal to, but I think my argument still holds (though I'm - according to the textbook - one out, so therefore my reasoning is somewhere wrong...);

Therefore \(\displaystyle P(X\geq\ x)\) implies that we have failed to get a group B sample x-1 times, therefore \(\displaystyle P(X\geq1)\) implies that we have failed to get a group B sample 0 times, which should equal 100% (you have argued using equals signs instead).


Now back to the first question - why the inequality has to change. You are correct, log_0.91 (0.002) is a positive value. So the reason lies elsewhere. One thing to notice is that the function 0.91^x has a negative slope, so as x gets bigger 0.91^x gets smaller, and eventually falls below the constant 0.002. But after taking the log of both sides the resulting LHS has an upward slope, whereas log(0.002) is still a constant, so as x gets bigger it eventually exceeds 0.002. The analogy I can think of is if you take the inverse of both sides of an inequality you have to switch the inequality. For example, given x > C if you take the inverse of bith sides you need to switch the inequality to get 1/x < 1/C.
However, your reasoning on this is solid, thanks :)

EDIT: here's the general equation that I have followed: \(\displaystyle P(X\geq\ x)=(1-p)^{x-1}\)
 
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