what is probability of getting J, Q, K, A from same suit out of 50 cards in three draws in first draw three cards are picked in second draw one card is picked and in third draw one card is picked

The problem asks to calculate the probability of drawing specific cards—J, Q, K, and A from the same suit—over three consecutive draws from a deck of 50 cards. The first draw takes 3 cards, the second draw takes 1 card, and the third draw takes 1 card.

### Assumptions:
- You want to get exactly **J, Q, K, and A** of the same suit.
- Cards are **not replaced** after being drawn.
- There are 4 suits, and for each suit, there are exactly 4 cards of interest (J, Q, K, A).

Let’s break it down step-by-step:

### 1. First Draw (3 cards):
There are 50 cards in the deck, and you need to pick 3 cards.

The probability of selecting 3 cards that are **all from the same suit and are either J, Q, or K** (i.e., three of the four specific cards from one suit) is:
- Out of 50 cards, you need to select these specific cards from the 4 cards (J, Q, K, A) of one suit.

For the first card, you have a \( \frac{4}{50} \) chance of picking one of the required cards from that suit.

After the first card is picked, 49 cards remain, and 3 of the required cards from that suit are left. So the probability for the second card is \( \frac{3}{49} \).

For the third card, you are left with 48 cards and 2 of the required cards, making the probability \( \frac{2}{48} \).

Thus, the probability of getting 3 specific cards (J, Q, K) from the same suit in the first draw is:

\[
P(\text{3 cards in first draw}) = \frac{4}{50} \times \frac{3}{49} \times \frac{2}{48}
\]

### 2. Second Draw (1 card):
Now, 47 cards remain in the deck, and only one card from the specific suit remains (the Ace). So, the probability of getting the Ace is:

\[
P(\text{Ace in second draw}) = \frac{1}{47}
\]

### 3. Third Draw (1 card):
After 4 cards have been drawn, 46 cards remain in the deck. Since you’ve already drawn all the required cards from that suit (J, Q, K, A), there is **no restriction** on this card. The probability of drawing **any card** here is simply 1:

\[
P(\text{any card in third draw}) = 1
\]

### Total Probability:
The total probability is the product of the probabilities from each step:

\[
P(\text{total}) = \left( \frac{4}{50} \times \frac{3}{49} \times \frac{2}{48} \right) \times \frac{1}{47} \times 1
\]

Now, let’s calculate this.

\[
P(\text{total}) = \frac{4 \times 3 \times 2}{50 \times 49 \times 48} \times \frac{1}{47}
\]

Let me compute this for you.The probability of drawing the Jack, Queen, King, and Ace from the same suit across three draws (where 3 cards are picked in the first draw, 1 card in the second draw, and 1 card in the third draw) is approximately \( 4.34 \times 10^{-6} \), or about **0.000434%**.

Would you like further clarification or more details on this calculation?

Here are five related questions to think about:
1. What happens to the probability if the deck size changes?
2. How would the probability change if you picked 2 cards in the second draw instead of 1?
3. How does the probability change if the draws are made with replacement?
4. What is the probability of getting only 3 of the 4 cards (J, Q, K, or A)?
5. How does the probability change if the cards are drawn in a different order?

**Tip:** In probability problems, always break down complex events into individual, independent steps, and multiply the probabilities of each step for the final outcome.

What is the probability of getting J, Q, K, A from the same suit out of 50 cards in three draws? In the first draw, three cards are picked, in the second draw, one card is picked, and in the third draw, one card is picked.

Calculate the probability of drawing the Jack, Queen, King, and Ace from the same suit in a 50-card deck over three consecutive draws. This problem uses combinatorics and probability rules. Learn the step-by-step process, perfect for advanced students interested in card game probabilities.

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