How do i find this statistics answer

[Biochemgirl2002]

Homework Statement

According to a report of the Nielson Company, 76% of Internet searches used Google
search engine. If 20 searches are randomly selected, find the probability that
more than 2 did not use Google.

Relevant Equations

nCr*0.76^n*0.24^r

i believe that the solution is to do 1- (less than 2 did use google)

so it would look like
1-
((20!/0!20!)0.76^0*0.24^20 +
(20!/1!19!)0.76^1*0.24^19 +
(20!/2!18!)0.76^2*0.24^18)
= 0.99
but logically, this does not make sense to be .99.
what am i doing wrong?

 

[Stephen Tashi]

How are you doing the arithmetic? It's difficult to get good answers in a calculation that involves quantities like $0.24^{20}$ unless you keep a lot of digits in the calculations.

To check your result, you can consult tables of the binomial distribution online. For example, the table for n = 20 given by http://www.sjsu.edu/people/saul.cohn/courses/stats/s0/BinomialProbabTable.pdf gives results for p = 0.20 and p = 0.25.
You can look at the values for 18,19 and 20 successes to get an idea of what should happen for p = 0.24.

Edit: There is this "binomial calculator" webpage https://www.anesi.com/binomial.htm?p=0.24&n=20
It displays a table if you select the "show full data table" option. (Of course, I haven't tried to check its answers.)

 

[Biochemgirl2002]

How are you doing the arithmetic? It's difficult to get good answers in a calculation that involves quantities like $0.24^{20}$ unless you keep a lot of digits in the calculations.

To check your result, you can consult tables of the binomial distribution online. For example, the table for n = 20 given by http://www.sjsu.edu/people/saul.cohn/courses/stats/s0/BinomialProbabTable.pdf gives results for p = 0.20 and p = 0.25.
You can look at the values for 18,19 and 20 successes to get an idea of what should happen for p = 0.24.

Edit: There is this "binomial calculator" webpage https://www.anesi.com/binomial.htm?p=0.24&n=20
It displays a table if you select the "show full data table" option. (Of course, I haven't tried to check its answers.)

On the table you provided me with, (the first link) the answer i got was correct with Probability of k successes or more. but i don't know if that is the right since it is asking the backwards answer for this question, since more than 2 did NOT use google, its tripping me up.

 

[LCKurtz]

Here's your calculation in Maple:

Edit, added: After looking more closely though I think you may want $$
\sum_{k=18}^{20} \binom {20} k p^k(1-p)^{20-k}$$ which comes out about $.108$.

 

[Stephen Tashi]

find the probability that more than 2 did not use Google.

i believe that the solution is to do 1- (less than 2 did use google)

The correct way to say it would be "1 - (2 or fewer did use google)"

That's what you calculated.

[Edit: No, it should be "1 - (2 or fewer did not use google)"]

 

[LCKurtz]

It's getting late and maybe I'm tired, but I think we are both wrong above, and it should be
$$\sum_{k=0}^{17} \binom {20} k p^k(1-p)^{20-k} = 1-\sum_{k=18}^{20} \binom {20} k p^k(1-p)^{20-k} $$ which is about $.891$.

 

[Biochemgirl2002]

It's getting late and maybe I'm tired, but I think we are both wrong above, and it should be
$$\sum_{k=0}^{17} \binom {20} k p^k(1-p)^{20-k} = 1-\sum_{k=18}^{20} \binom {20} k p^k(1-p)^{20-k} $$ which is about $.891$.

alright, the p is different than any variable I've used in this course, is p always going to be 0.76? and k is reducing from 20 until 2?

 

[LCKurtz]

As you gave and I put in the Maple worksheet $p=.76,~ q = 1-p = .24$

 

[Biochemgirl2002]

As you gave and I put in the Maple worksheet $p=.76,~ q = 1-p = .24$

thank you so much for taking time out of your day to help me out! i appreciate it