What is the probability that they form a match? On a dating site, users can select $5$ out of $24$ adjectives to describe themselves. A match is declared between two users if they match on at least $4$ adjectives. If Alice and Bob randomly pick adjectives, what is the probability that they form a match?
My Attempt
Probability space is $24\choose 5$$^2$ we have match for $4$ or $5$ so
for $5$ it's just $24\choose 5$ and for for it's $24\choose 4$ so the overall probability is
$$\frac{^{24}C_5+^{24}C_4}{(^{24}C_5)^2}$$
the given answer is $$\frac{^{24}C_5.(1+5(24-5))}{^{24}C_5.^{24}C_5}$$
 A: Going by your method, we count all cases where i) exactly $4$ adjectives match and ii) all $5$ adjectives match.
You are right that number of ways for $5$ match is ${24 \choose 5}$.
Now number of ways for exactly $4$ adjective match -
${24 \choose 4} \cdot 20 \cdot 19$
We first choose $4$ adjectives that are match and then there are $20 \cdot 19$ ways for both of them to choose an adjective each from remaining $20$ adjectives. You missed the last part.
So desired probability is,
$\displaystyle \cfrac{{24 \choose 5} + {24 \choose 4} \cdot 20 \cdot 19} {{24 \choose 5} \cdot {24 \choose 5}}$
which is same as the given answer.
A: For case 4, we can see this.
Alice chooses 5 adjectives, then there are $24 \choose 5$ posibilities.
Then Bob can choose $5 \choose 4$ = $5$  ways for 4 adjective which are same as Alice. The last adjective Bob can choose has $24-5 \choose 1$ = $24-5$ posibilities because last adjective is not in 5 adjectives Alice chosen.
Thus, Bos has $24\choose 5$5(24-5) way to choose adjectives which has 4 same adjectives with Alice.
So the answer is same as given.
