Question

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.

Explanation:

1. For total possible outcomes of choosing 5 cards from 52 (order doesn't matter in a hand), we use combinations (\(_{n}C_{k}\)). So \(_{52}C_{5}\) is correct, \(_{52}P_{5}\) (permutations, order matters) is wrong.
2. Probability of 2 diamonds (13 diamonds, choose 2) and 3 hearts (13 hearts, choose 3): \(\frac{_{13}C_{2}\times_{13}C_{3}}{_{52}C_{5}}\). Calculate \(_{13}C_{2}=\frac{13!}{2!11!}=78\), \(_{13}C_{3}=\frac{13!}{3!10!}=286\), \(_{52}C_{5}=\frac{52!}{5!47!}=2598960\). Numerator: \(78\times286 = 22308\). Probability: \(\frac{22308}{2598960}\approx0.00858\), not 0.089.
3. Probability of 5 spades: \(_{13}C_{5}/_{52}C_{5}\). \(_{13}C_{5}=\frac{13!}{5!8!}=1287\). Probability: \(\frac{1287}{2598960}\approx0.000495\), not 0.05.
4. Probability of 5 clubs: same as 5 spades, \(\frac{_{13}C_{5}}{_{52}C_{5}}\approx\frac{1287}{2598960}\approx0.0005\) (approximate). Also, the first statement (using \(_{52}C_{5}\)) is correct, and the last statement (five clubs probability ~0.0005) is correct. Wait, re - check:

• First statement: Choosing 5 - card hand, order doesn't matter, so combinations \(_{52}C_{5}\) is correct.
• Fifth statement: \(_{13}C_{5}=\frac{13\times12\times11\times10\times9}{5\times4\times3\times2\times1}=1287\), \(_{52}C_{5}=2598960\), \(1287\div2598960\approx0.000495\approx0.0005\), correct.
• Let's re - check the second statement: Permutations are for ordered arrangements, in a card hand, the order of the 5 cards doesn't matter, so \(_{52}P_{5}\) is wrong.
• Third statement: \(_{13}C_{2}=78\), \(_{13}C_{3}=286\), \(78\times286 = 22308\), \(22308\div2598960\approx0.00858

eq0.089\), wrong.

• Fourth statement: \(1287\div2598960\approx0.000495\approx0.0005\), not 0.05, wrong.
• So the true statements are: "The total possible outcomes can be found using \(_{52}C_{5}\)" and "The probability of choosing five clubs is roughly 0.0005". Wait, but let's check the five clubs probability again. \(_{13}C_{5}=\frac{13!}{5!(13 - 5)!}=\frac{13\times12\times11\times10\times9}{5\times4\times3\times2\times1}=1287\), \(_{52}C_{5}=\frac{52!}{5!(52 - 5)!}=\frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1}=2598960\). Then \(1287\div2598960\approx0.000495\approx0.0005\), correct. And the total outcomes: since we are choosing a set of 5 cards (order doesn't matter), combinations \(_{52}C_{5}\) is correct. The permutation \(_{52}P_{5}\) is for ordered sequences, so wrong. The two diamonds and three hearts: \(_{13}C_{2}\times_{13}C_{3}=78\times286 = 22308\), \(22308\div2598960\approx0.00858\), not 0.089. Five spades: same as five clubs, ~0.0005, not 0.05. 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.