probability, binomial experiment

Question

According to a report of the Nielsen company 65% of internet searches in may 2010 used the google search engine. assume that a sample of 20 searches is studied. 

 

a) what is the probability that exactly 16 of them used google?

b) what is the probability that more than 16 used google?

c) what is the probability that between 15 and 18, inclusive used google?

Answer 1

In a binomial probability problem, the formula for the probability of N tries occurring with a specific result, out of M total tries is:

 

_MC_N P(True)^N P(False)^M-N

 

_NC_{M =} M! / (N! (M-N)!

P(True) is the probability that the specific result occurs in each try

P(False) is the probability that the specific result does not occur, in each try

 

P(False) = 1 - P(True), so this can be written as:

 

_MC_N P(True)^N (1 - P(True)^M-N)

 

So...

 

a.  _MC_N P(True)^N (1 - P(True)^M-N) = [20! / (16! 4!)] * .65¹⁶ * .35⁴ = 0.0738

 

b. You have to do this calculation for all N > 16 (17, 18, 19, and 20), and add them:

 

[20! / (17! 3!)] * .65¹⁷ * .35³ +

[20! / (18! 2!)] * .65¹⁸ * .35² +

[20! / (19! 1!)] * .65¹⁹ * .35¹ +

[20! / (20! 0!)] * .65²⁰ * .35⁰ =

 

0.0323 +  0.0100 + .0020 + .0002 = 0.0445

 

c. You have to do this calculation for N = 15, 16, 17, and 18, and add them:

 

[20! / (15! 5!)] * .65¹⁵ * .35⁵ +

[20! / (16! 4!)] * .65¹⁶ * .35⁴ +

[20! / (17! 3!)] * .65¹⁷ * .35³ +

[20! / (18! 2!)] * .65¹⁸ * .35² =

 

0.1272 + 0.0738 + 0.0323 +  0.0100 = 0.2433

 

Check my calculations!