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,
\(P(X\geq2)=0.765\)
