Xrfgtjtk

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

What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases,
then we need to find a conditional probability
$P(Xgeq 2 | Y)=P(X= 2,3,text{ or }4 | Y)=frac{P({Xgeq 2} cap Y)}{P(Y)}$
but then I don't know how to continue?

Any help...

You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.

If $N$ is the number of dictionaries where the keyword is found then $$Pr(N=n) = frac{{5 choose n}{4 choose 4-n}}{9 choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.

So for $n=0,1,2,3,4$ this gives probabilities $frac{1}{126},frac{20}{126},frac{60}{126},frac{40}{126},frac{5}{126}$.

Add up the last three (and change to lowest terms) and you have your solution.

Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.

$$text{ Henry: } P(X = x) = frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}}; text{ me: } P(X = x) = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$

But

$$frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}} = frac{M!}{x! (M-x)!} frac{(N-M)!}{(K-x)! (N-M-K+x)!} frac{K! (N-K)!}{N!}$$
$$= frac{K!}{x! (K-x)!} frac{(N-K)!}{(M-x)! (N-K-M+x)!} frac{M! (N-M)!}{N!} = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$

Thus the two answers give the same result.

(Original answer.)

I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X geq 2)$.

$$P(X = 2) = frac{binom{4}{2} binom{5}{3}}{binom{9}{5}}$$ because there are $binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.

I'll let you calculate $P(X = 3)$ and $P(X = 4)$.

(FYI, $X$ has what's called a hypergeometric distribution.)

Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword.

Let $c=4$ databases are chosen and $s$ databases have the keyword.

The question is asking to find $sge 2$, i.e. $s=color{red}2,color{green}3,color{blue}4$.

There are ${tchoose c}={9choose 4}$ ways to choose $c$ databases from total $t$ databases.

Case 1: There are ${kchoose 2}={5choose color{red}2}$ ways to choose from $k$ and ${nchoose 2}={4choose 2}$ ways from $n$, hence: ${5choose 2}{4choose 2}$.

Case 2: There are ${kchoose 3}={5choose color{green}3}$ ways to choose from $k$ and ${nchoose 1}={4choose 1}$ ways from $n$, hence: ${5choose 3}{4choose 1}$.

Case 3: There are ${kchoose 4}={5choose color{blue}4}$ ways to choose from $k$ and ${nchoose 0}={4choose 0}$ ways from $n$, hence: ${5choose 4}{4choose 0}$.

Hence, the required probability is:
$$P(sge 2)=P(s=2)+P(s=3)+P(s=4)=\
frac{{5choose 2}{4choose 2}}{9choose 4}+frac{{5choose 3}{4choose 1}}{9choose 4}+frac{{5choose 4}{4choose 0}}{9choose 4}=\
frac{60+40+5}{126}=frac{35}{42}approx 0.83.$$
Note that you can also use the complement, i.e.:
$$P(sge 2)=1-P(s=0)-P(s=1).$$

Thus, in general, the formula is:
$$P(sge r)=sum_{i=r}^c P(s=i)=sum_{i=r}^c frac{{kchoose i}{nchoose c-i}}{{tchoose c}}.$$

Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.