5 cards are drawn from well shuffled deck of cards (without replacement) .What is the probability that three of them are between 5 to 7? probability probability-theory combinations If the question asked about only 3 of them are between 5 to 7 then the answer is easy it will be $$ \frac{\binom{12}{3}\binom{40}{2}}{\binom{52}{5}}. $$ But here it is different I thought of the following two answers
$12/52 * 11/51 * 10/50 *1 =11/1105$ Since it doesn't matter what other 2 cards are but since the order doesn't matter we must divided the above by 3! So the answer = 11/6630 which equal to $12C3/(52 * 51 *50)$
$(12C3*49C2)/52C5 =22/221$ because I think the answer must has 52C5 as it denominator.
As far as i have understood , you want to say three of five cards are in $[5-7]$ and two of five cards are not in $[5-7]$. This means that $\color{blue}{exactly}$ three cards in $[5-7]$.
Lets call the cards between $[5-7]$ as $\color{red}{A}$ and the others as $\color{green}{B}$. To determine their selection order , we should arrange $\color{red}{A,A,A}$$\color{green}{,B,B}$ . We can do it by $$\frac{5!}{2! \times 3!}=10$$
Moreover , question say that no replacements . Then ,we know that there are $12$ different cands in $[5-7]$. We can assign them to $\color{red}{A,A,A}$ by $P(12,3)$ ways . The rest is done by $P(40,2)$ ways.
The denominator will be $P(52,5)$ instead of $C(52,5)$ because there is not any replacement and when we select each card , the total number of cards will be selected are going to decrease by one by.
CASE II-) If there are at least $3$ cards in $[5-7]$ , then :
Subcase I-)Lets call the cards between $[5-7]$ as $\color{red}{A}$ and the others as $\color{green}{B}$. To determine their selection order , we should arrange $\color{red}{A,A,A}$$\color{green}{,B,B}$ . We can do it by $$\frac{5!}{2! \times 3!}=10$$
Moreover , question say that no replacements . Then ,we know that there are $12$ different cands in $[5-7]$. We can assign them to $\color{red}{A,A,A}$ by $P(12,3)$ ways . The rest is done by $P(40,2)$ ways.
The denominator will be $P(52,5)$ instead of $C(52,5)$ because there is not any replacement and when we select each card , the total number of cards will be selected are going to decrease by one by.
Subcase II-) There are $4$ desired card and $1$ undesired card such that :
Lets call the cards between $[5-7]$ as $\color{red}{A}$ and the others as $\color{green}{B}$. To determine their selection order , we should arrange $\color{red}{A,A,A,A}$$\color{green}{,B}$ . We can do it by $$\frac{5!}{1! \times 4!}=5$$
We can assign them to $\color{red}{A,A,A,A}$ by $P(12,4)$ ways . The rest is done by $P(40,1)$ ways.
The denominator will be $P(52,5)$ instead of $C(52,5)$ because there is not any replacement and when we select each card , the total number of cards will be selected are going to decrease by one by.
Subcase III-)Lets call the cards between $[5-7]$ as $\color{red}{A}$ and the others as $\color{green}{B}$. To determine their selection order , we should arrange $\color{red}{A,A,A,A,A}$ . We can do it by $$\frac{5!}{5!}=1$$
Moreover , question say that no replacements . Then ,we know that there are $12$ different cands in $[5-7]$. We can assign them to $\color{red}{A,A,A,A,A}$ by $P(12,5)$ ways .
The denominator will be $P(52,5)$ instead of $C(52,5)$ because there is not any replacement and when we select each card , the total number of cards will be selected are going to decrease by one by.
$\color{blue}{RESULT=}$ $$(\frac{5!}{2! \times 3!} \times \frac {P(12,3) \times P(40,2)}{P(52,5)}) + (\frac{5!}{1! \times 4!} \times \frac {P(12,4) \times P(40,1)}{P(52,5)}) + (\frac{5!}{5!} \times \frac {P(12,5) }{P(52,5)}) = 0.0739495... $$