To find the probability of each event, we need to determine the total number of possible outcomes and the number of favorable outcomes for each event.

(a) Probability of 4 aces:
There are 4 aces in a deck of 52 cards. To draw 4 aces, we need to choose 4 cards out of the 4 aces, which can be done in 4C4 ways. The total number of possible outcomes is choosing any 5 cards out of 52, which can be done in 52C5 ways. Therefore, the probability of drawing 4 aces is 4C4 / 52C5.

(b) Probability of 4 aces and 1 king:
There are 4 aces and 4 kings in a deck of 52 cards. To draw 4 aces and 1 king, we need to choose 4 cards out of the 4 aces and 1 card out of the 4 kings. This can be done in 4C4 * 4C1 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of drawing 4 aces and 1 king is (4C4 * 4C1) / 52C5.

(c) Probability of 3 tens and 2 jacks:
There are 4 tens and 4 jacks in a deck of 52 cards. To draw 3 tens and 2 jacks, we need to choose 3 cards out of the 4 tens and 2 cards out of the 4 jacks. This can be done in 4C3 * 4C2 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of drawing 3 tens and 2 jacks is (4C3 * 4C2) / 52C5.

(d) Probability of obtaining a nine, ten, jack, queen, king in any order:
There are 4 cards each of nine, ten, jack, queen, and king in a deck of 52 cards. To obtain these cards in any order, we need to choose 1 card out of the 4 nines, 1 card out of the 4 tens, 1 card out of the 4 jacks, 1 card out of the 4 queens, and 1 card out of the 4 kings. This can be done in 4C1 * 4C1 * 4C1 * 4C1 * 4C1 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of obtaining a nine, ten, jack, queen, king in any order is (4C1 * 4C1 * 4C1 * 4C1 * 4C1) / 52C5.

(e) Probability of 3 cards of one suit and 2 cards of another suit:
There are 13 cards of each suit in a deck of 52 cards. To have 3 cards of one suit and 2 cards of another suit, we need to choose 3 cards out of the 13 cards of one suit and 2 cards out of the 13 cards of another suit. This can be done in 13C3 * 13C2 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of having 3 cards of one suit and 2 cards of another suit is (13C3 * 13C2) / 52C5.

(f) Probability of obtaining at least one ace:
To find the probability of obtaining at least one ace, we can find the probability of not obtaining any ace and subtract it from 1. The probability of not obtaining any ace is the same as the probability of drawing 5 cards from the 48 non-ace cards in the deck, which can be calculated as 48C5 / 52C5. Therefore, the probability of obtaining at least one ace is 1 - (48C5 / 52C5).

Please note that these calculations assume that the cards are well-shuffled and drawn randomly.

Question

To find the probability of each event, we need to determine the total number of possible outcomes and the number of favorable outcomes for each event.

(a) Probability of 4 aces:
There are 4 aces in a deck of 52 cards. To draw 4 aces, we need to choose 4 cards out of the 4 aces, which can be done in 4C4 ways. The total number of possible outcomes is choosing any 5 cards out of 52, which can be done in 52C5 ways. Therefore, the probability of drawing 4 aces is 4C4 / 52C5.

(b) Probability of 4 aces and 1 king:
There are 4 aces and 4 kings in a deck of 52 cards. To draw 4 aces and 1 king, we need to choose 4 cards out of the 4 aces and 1 card out of the 4 kings. This can be done in 4C4 * 4C1 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of drawing 4 aces and 1 king is (4C4 * 4C1) / 52C5.

(c) Probability of 3 tens and 2 jacks:
There are 4 tens and 4 jacks in a deck of 52 cards. To draw 3 tens and 2 jacks, we need to choose 3 cards out of the 4 tens and 2 cards out of the 4 jacks. This can be done in 4C3 * 4C2 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of drawing 3 tens and 2 jacks is (4C3 * 4C2) / 52C5.

(d) Probability of obtaining a nine, ten, jack, queen, king in any order:
There are 4 cards each of nine, ten, jack, queen, and king in a deck of 52 cards. To obtain these cards in any order, we need to choose 1 card out of the 4 nines, 1 card out of the 4 tens, 1 card out of the 4 jacks, 1 card out of the 4 queens, and 1 card out of the 4 kings. This can be done in 4C1 * 4C1 * 4C1 * 4C1 * 4C1 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of obtaining a nine, ten, jack, queen, king in any order is (4C1 * 4C1 * 4C1 * 4C1 * 4C1) / 52C5.

(e) Probability of 3 cards of one suit and 2 cards of another suit:
There are 13 cards of each suit in a deck of 52 cards. To have 3 cards of one suit and 2 cards of another suit, we need to choose 3 cards out of the 13 cards of one suit and 2 cards out of the 13 cards of another suit. This can be done in 13C3 * 13C2 ways. The total number of possible outcomes is still 52C5. Therefore, the probability of having 3 cards of one suit and 2 cards of another suit is (13C3 * 13C2) / 52C5.

(f) Probability of obtaining at least one ace:
To find the probability of obtaining at least one ace, we can find the probability of not obtaining any ace and subtract it from 1. The probability of not obtaining any ace is the same as the probability of drawing 5 cards from the 48 non-ace cards in the deck, which can be calculated as 48C5 / 52C5. Therefore, the probability of obtaining at least one ace is 1 - (48C5 / 52C5).

Please note that these calculations assume that the cards are well-shuffled and drawn randomly.

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