Fh. Five cards are drawn from a standard deck without replacement. What is the probability of that they will be a 10 , jack, queen, king and ace, in that order?

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To find the probability of drawing a 10, jack, queen, king, and ace in that specific order from a standard deck of cards without replacement, we can follow these steps:

1. **Identify the total number of cards in the deck**: A standard deck has 52 cards.

2. **Calculate the probability of drawing each card in the specified order**:
   - The probability of drawing a 10 first: There are 4 tens in the deck, so the probability is $$\frac{4}{52}$$.
   - The probability of drawing a jack second: After drawing a 10, there are 51 cards left, and still 4 jacks, so the probability is $$\frac{4}{51}$$.
   - The probability of drawing a queen third: After drawing a 10 and a jack, there are 50 cards left, and still 4 queens, so the probability is $$\frac{4}{50}$$.
   - The probability of drawing a king fourth: After drawing a 10, jack, and queen, there are 49 cards left, and still 4 kings, so the probability is $$\frac{4}{49}$$.
   - The probability of drawing an ace fifth: After drawing a 10, jack, queen, and king, there are 48 cards left, and still 4 aces, so the probability is $$\frac{4}{48}$$.

3. **Multiply the probabilities together**: The overall probability of drawing these cards in that specific order is the product of the individual probabilities.

Now, let's calculate this step by step:

$$
P = \frac{4}{52} \times \frac{4}{51} \times \frac{4}{50} \times \frac{4}{49} \times \frac{4}{48}
$$

Now, I will calculate this expression.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(\frac{4}{52}\right)\times \frac{4}{51}\left(\frac{4}{50}\right)\times \frac{4}{49}\left(\frac{4}{48}\right)$$
- step1: Reduce the fraction:
  $$\frac{1}{13}\times \frac{4}{51}\left(\frac{4}{50}\right)\times \frac{4}{49}\left(\frac{4}{48}\right)$$
- step2: Reduce the fraction:
  $$\frac{1}{13}\times \frac{4}{51}\times \frac{2}{25}\times \frac{4}{49}\left(\frac{4}{48}\right)$$
- step3: Reduce the fraction:
  $$\frac{1}{13}\times \frac{4}{51}\times \frac{2}{25}\times \frac{4}{49}\times \frac{1}{12}$$
- step4: Multiply the terms:
  $$\frac{4}{663}\times \frac{2}{25}\times \frac{4}{49}\times \frac{1}{12}$$
- step5: Multiply the terms:
  $$\frac{8}{16575}\times \frac{4}{49}\times \frac{1}{12}$$
- step6: Multiply the terms:
  $$\frac{32}{812175}\times \frac{1}{12}$$
- step7: Reduce the numbers:
  $$\frac{8}{812175}\times \frac{1}{3}$$
- step8: Multiply the fractions:
  $$\frac{8}{812175\times 3}$$
- step9: Multiply:
  $$\frac{8}{2436525}$$
The probability of drawing a 10, jack, queen, king, and ace in that specific order from a standard deck of cards without replacement is:

$$
P = \frac{8}{2436525} \approx 3.283365 \times 10^{-6}
$$

This means that the probability is approximately $$0.000003283365$$, which is a very small probability, indicating that this specific sequence of draws is quite unlikely.
The probability of drawing a 10, jack, queen, king, and ace in that specific order from a standard deck without replacement is approximately $$3.28 \times 10^{-6}$$ or 0.00000328.

Fh. Five cards are drawn from a standard deck without replacement. What is the probability of that they will be a 10 , jack, queen, king and ace, in that order?

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