a deck of playing cards has four suits, with thirteen cards in each suit consisting of the numbers 2 through 10, a jack, a queen, a king, and an ace. the four suits are hearts, diamonds, spades, and clubs. a hand of five cards will be chosen at random. which statements are true? check all that apply. the total possible outcomes can be found using $_{52}c_5$. the total possible outcomes can be found using $_{52}p_5$. the probability of choosing two diamonds and three hearts is 0.089. the probability of choosing five spades is roughly 0.05 the probability of choosing five clubs is roughly 0.0005.

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# Brief Explanations:
1. **For $_{52}C_5$ and $_{52}P_5$**:
   - When choosing a hand of five cards, the order of the cards in the hand does not matter. The combination formula $_{n}C_{r}=\frac{n!}{r!(n - r)!}$ is used for unordered selections, and the permutation formula $_{n}P_{r}=\frac{n!}{(n - r)!}$ is used for ordered selections. Since a hand of cards is an unordered set, the total number of possible 5 - card hands from a 52 - card deck is given by $_{52}C_5$, not $_{52}P_5$. So the first statement is true and the second is false.
2. **Probability of two diamonds and three hearts**:
   - The number of ways to choose 2 diamonds from 13 diamonds is $_{13}C_2=\frac{13!}{2!(13 - 2)!}=\frac{13\times12}{2\times1}=78$.
   - The number of ways to choose 3 hearts from 13 hearts is $_{13}C_3=\frac{13!}{3!(13 - 3)!}=\frac{13\times12\times11}{3\times2\times1}=286$.
   - The number of ways to choose 5 cards from 52 cards is $_{52}C_5=\frac{52!}{5!(52 - 5)!}=\frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1}=2598960$.
   - The number of favorable outcomes (2 diamonds and 3 hearts) is $_{13}C_2\times_{13}C_3=78\times286 = 22308$.
   - The probability $P=\frac{22308}{2598960}\approx0.00858$? Wait, no, wait, I made a mistake. Wait, $_{13}C_2=\frac{13!}{2!11!}=\frac{13\times12}{2}=78$, $_{13}C_3=\frac{13!}{3!10!}=\frac{13\times12\times11}{6}=286$, and $78\times286 = 22308$. And $_{52}C_5 = 2598960$. Then $\frac{22308}{2598960}\approx0.00858$? But the statement says 0.089. Wait, maybe I miscalculated. Wait, no, maybe the user made a typo or I messed up. Wait, no, let's recalculate: $_{13}C_2 = 78$, $_{13}C_3=286$, $78\times286 = 22308$. $22308\div2598960\approx0.00858$, which is not 0.089. Wait, maybe I made a mistake in the combination. Wait, no, maybe the problem is about two diamonds and three hearts, but maybe I should consider that the total number of ways to choose 2 diamonds and 3 hearts is $_{13}C_2\times_{13}C_3$, and the total number of 5 - card hands is $_{52}C_5$. Wait, maybe the correct calculation is: $_{13}C_2=\frac{13\times12}{2}=78$, $_{13}C_3=\frac{13\times12\times11}{6}=286$, $78\times286 = 22308$, $22308\div2598960\approx0.00858$, so this statement is false? Wait, maybe I made a mistake. Wait, no, let's check again. Wait, the formula for probability is $\frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$. The number of favorable outcomes for 2 diamonds and 3 hearts is $_{13}C_2\times_{13}C_3$, and total outcomes is $_{52}C_5$. So $_{13}C_2 = 78$, $_{13}C_3 = 286$, product is 22308. $_{52}C_5=2598960$. $22308\div2598960\approx0.00858$, so the third statement is false.
3. **Probability of five spades**:
   - The number of ways to choose 5 spades from 13 spades is $_{13}C_5=\frac{13!}{5!(13 - 5)!}=\frac{13\times12\times11\times10\times9}{5\times4\times3\times2\times1}=1287$.
   - The probability $P=\frac{_{13}C_5}{_{52}C_5}=\frac{1287}{2598960}\approx0.000495\approx0.0005$, not 0.05. So the fourth statement is false.
4. **Probability of five clubs**:
   - The number of ways to choose 5 clubs from 13 clubs is $_{13}C_5 = 1287$ (same as five spades, since there are 13 clubs).
   - The probability $P=\frac{_{13}C_5}{_{52}C_5}=\frac{1287}{2598960}\approx0.000495\approx0.0005$. So the fifth statement is true. Also, the first statement ($_{52}C_5$ for total outcomes) is true. Wait, let's re - evaluate the third statement. Wait, maybe I made a mistake in the third statement. Wait, two diamonds and three hearts: number of diamonds is 13, number of hearts is 13. So $_{13}C_2=\frac{13!}{2!11!}=78$, $_{13}C_3=\frac{13!}{3!10!}=286$, $78\times286 = 22308$. $_{52}C_5 = 2598960$. $22308\div2598960\approx0.00858$, which is about 0.0086, not 0.089. So the third statement is false. Wait, but maybe the question has a different approach. Wait, maybe the user made a mistake, but according to the calculations:
     - Statement 1: True (since combination is for unordered, which is correct for a hand of cards).
     - Statement 2: False (permutation is for ordered, not for a hand).
     - Statement 3: False (calculated probability is ~0.0086, not 0.089).
     - Statement 4: False (calculated probability is ~0.0005, not 0.05).
     - Statement 5: True (calculated probability is ~0.0005).

Wait, but let's recalculate the probability of five clubs: $_{13}C_5=\frac{13\times12\times11\times10\times9}{5\times4\times3\times2\times1}=1287$, $_{52}C_5 = 2598960$, $1287\div2598960\approx0.000495\approx0.0005$, so statement 5 is true. And statement 1 is true. Wait, maybe I made a mistake in the third statement. Let's check again: two diamonds and three hearts. The number of diamonds is 13, hearts is 13. So the number of ways to choose 2 diamonds: $C(13,2)=78$, 3 hearts: $C(13,3)=286$. The number of favorable is $78\times286 = 22308$. Total number of 5 - card hands: $C(52,5)=2598960$. $22308\div2598960 = 0.00858\approx0.0086$, which is not 0.089. So statement 3 is false.

So the true statements are:
- The total possible outcomes can be found using $_{52}C_5$.
- The probability of choosing five clubs is roughly 0.0005.

# Answer:
- The total possible outcomes can be found using $_{52}C_5$.
- The probability of choosing five clubs is roughly 0.0005.

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