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The deck is composed of cards from 6 to A:

6 7 8 9 10 J Q K A of Diamonds

6 7 8 9 10 J Q K A of Hearts

6 7 8 9 10 J Q K A of Spades

6 7 8 9 10 J Q K A of Clubs

  • The deck is shuffled and I draw 05 cards.

Say I drew the cards: 9 9 9 J A

A1) If I exchange the J and A and draw more two cards, what's the probability of one of these new 02 cards being a 9?

A2) If I exchange the J and A and draw more two cards, what's the probability of drawing a pair?

B1) If I exchange only the the J and draw one more card, what's the probability of this new card being a J?

B2) If I exchange only the the J and draw one more card, what's the probability of this new card being a 9?

(I presume you see this is a Poker questions, but the root of the questions is about probability, that's why I'm posting it here.)

Thank you very much.

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I find the most general and useful approach in solving these types of probability questions is to count the # of desired results and divide by the total # of possibilities. That is, $Probability =\dfrac{\# desired}{\# total}$.

Here's a breakdown of the steps for A1.

Assuming you "discard" (without replacement) the J and A, we had 36 cards total, and drew 5, so we have 31 cards left in the deck, and 1 of those is a 9.

We start by counting how many desired results there are. And represent the two cards by "_ _".

$\textbf{Step 1: Count the positions.}$

_ _

We just need a 9 to be one of the 2 cards, so order doesn't matter. We had 2 spots to choose from, so at this point we have Desired = 2

$\textbf{Step 2: Fix a card.}$

9 _

There's only 1 card that is desired for that spot. Desired = $2 \times 1$

$\textbf{Step 3: Fill rest of positions}$

9 A

We can have any other card for the other position, and there are 30 cards left. Here we have picked "A" arbitrarily to represent any card. Desired = $2 \times 1 \times 30 = 60$

$\textbf{Step 4: Count total possibilities.}$

_ _

We have 31 choices for the first card, and a 30 choices for the second card. Thus, Total = $31 \times 30$.

Thus, $\dfrac{\# desired}{\# total} = \dfrac{2\times 1 \times 30}{31\times 30} = \dfrac{2}{31}$

The method for solving the rest is highly similar, so that should get you started.

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