Draw 3 card simultaneously from a standard deck without replacement. I have this the following question with 4 parts, but I don't know how the numbers are calculated for parts 1,3,4. Can anyone help me.
We simulteneously draw three cards:
A-) Probability of observing exactly 2 diffrent shapes: ANSWER: 0.551-->HOW
B-) Probability of Observing at least one king and one queen. --> I know how to do this part.
C-) Probability of observing 2 jacks and diamond as the third card. -->ANSWER: 0.0034 -->HOW
D-)Probability of observing 3 different face value. --> ANSWER: 0.83  --> NO IDEA HOW
For part A, I remember there was a method with combinations but I don't remember the method. Does anyone know how?
I tried 39/51 but it is not correct. I thought 1st could be anything 2nd should be one of the remaining 39 cards from shapes that are not the same as the shape of first.
For C I did:
(4x3x12)/(52x51x50) + (3x2x13)/(52x51x50) --> 2 jacks including  diamond + 2 jack no diamond
But the answer is 4 times the number I found. Where does *4 coming from.
 A: *

*A. Probability of only picking one suit is $4\cdot\frac{13\cdot12\cdot11}{52\cdot51\cdot50}$. Probability of picking $3$ suits is $1\cdot\frac34\cdot\frac12$. You can finish from here


*C. Case 1: The last card is a jack: Then, there is one (non-diamond) jack left to pick, and two possible places it could be picked. So, we have $2\cdot\frac{48}{52}\cdot\frac3{51}\cdot\frac1{50}$. Case 2: No jack of diamonds is picked: the first two cards must be jacks. So, we have $\frac3{52}\cdot\frac2{51}\cdot\frac{13}{50}$. Case 3: A jack of diamonds is picked, but not as the last card: $2\cdot\frac3{52}\cdot\frac1{51}\cdot\frac{12}{52}$. The sum of these cases is the answer.


*D. $1\cdot\frac{48}{51}\cdot\frac{44}{50}$
A: For A)
Exactly 2 shapes mean 2 out of the 3 cards is of the same suit. The number of ways we can get 2 same cards out of any suit is ${13 \choose 2}$. After the 2 cards, we can choose any cards from any suit, and there are 39 of them. And since we have 4 suits, we multiply by $4$.
Diving by the total number of ways we can get 3 cards, we get
$\frac{4({13 \choose 2} 39)}{{52 \choose 3}}$
A: I'll help you out with the first part, and then hopefully you will be able to try a similar process for the rest.  To get the probability of exactly two of the three drawn cards having different suits I'm going to think about the situation as follows.  There are two possibilities:

*

*Draw a card, now there are $51$ cards left.

*The second card is the same suit as the first: probability $\frac{12}{51}$, $50$ cards left.

*The third card is a different suit from the first two: probability $\frac{39}{50}$
or

*

*Draw a card, now there are $51$ cards left

*The second card is a different suit from the first: probability $\frac{39}{51}$, $50$ cards left.

*The third card is either of the first two suits, of which there are $12$ left in the deck each: probability $\frac{24}{50}$
Probability for the first option:$$\frac{12}{51} \cdot \frac{39}{50}.$$
Probability for the second option:$$\frac{39}{51} \cdot \frac{24}{50}.$$
Total probability that either event occurs:$$\frac{12}{51} \cdot \frac{39}{50} + \frac{39}{51} \cdot \frac{24}{50} = \frac{234}{425} \approx 0.551.$$
