Question

An internet search engine looks for a keyword in 9 databases, searching them in a random order. Only 5 of these databases contain the given keyword. Find the probability that it will be found in at least 2 of the first 4 searched databases.

Answer 1

I would guess 3 5 7 9 are wrong just because

Answer 2

Hence,

`P(X\geq2)=0.765`

Explanation:

There are total 9 databases.

Only 5 of these databases contain the given keyword.

we have to find the probability that it will be found in at least 2 of the first 4 searched databases.

So, we will use the binomial to find this probability.

We know that the probability of r successes out of the n outcomes is calculated as:

`P(X=r)=n_C_rp^r(1-p)^(n-r)`

Where p denote the probability of success.

and

`n_C_r=(n!)/(r!* (n-r)!)`

p=5/9 ( since out of the 9 databases 5 contain the given keyword)

1-p=4/9.

so, we have to find:

`P(X\geq2)`

Here n=4

Also,

`P(X\geq2)=P(X=2)+P(X=3)+P(X=4)`

`P(X=2)=4_C_2* ((5)/(9))^2* ((4)/(9))^2----------(1)`

`P(X=3)=4_C_3* ((5)/(9))^3* ((4)/(9))^1----------(2)`

`P(X=4)=4_C_4* ((5)/(9))^4* ((4)/(9))^0---------(3)`

Hence, the probability is calculated by adding equation (1),(2) and (3).

Hence,

`P(X\geq2)=0.765`