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

Answer:

Hence,

$P(X\geq2)=0.765$

Step-by-step 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

$n_C_r=\dfrac{n!}{r!\times (n-r)!}$

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,

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

$P(X=2)=4_C_2\times (\dfrac{5}{9})^2\times (\dfrac{4}{9})^2----------(1)$

$P(X=3)=4_C_3\times (\dfrac{5}{9})^3\times (\dfrac{4}{9})^1----------(2)$

$P(X=4)=4_C_4\times (\dfrac{5}{9})^4\times (\dfrac{4}{9})^0---------(3)$

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

Hence,

$P(X\geq2)=0.765$

Answer 2

I would guess 3 5 7 9 are wrong just because