A deck of playing cards contains 52 cards, four of which are aces. What is the probability that the deal of a five-card hand provides:
a. What is the appropriate probability distribution for X? Explain how X satisfies the properties of the distribution.
b. A pair of Aces?
c. Exactly one ace?
d. No aces?
e. At least one ace?
a. I don't know what X signifies.
b. First find the probability of two aces, then 3 non-aces.
(4/52)(3/51)(48/50)(47/49)(46/48) = ?
c, d. Use similar to process above.
e. At least one ace = 1, 2, 3, or 4 aces. Use answers from b and c, plus calculate probability of 3 aces and 4 aces.
The probability of either-or = sum of individual probabilities.
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To answer these questions, we need to understand the concept of probability distributions and how they relate to the given scenario with the deck of cards.
a. Probability Distribution for X:
In this scenario, X represents the number of aces in a five-card hand drawn from a standard deck of playing cards. The appropriate probability distribution for X is the binomial distribution, as it satisfies the properties required for this type of distribution.
The properties satisfied by the binomial distribution are as follows:
1. There are a fixed number of trials: In this case, five cards are being drawn.
2. Each trial has two possible outcomes: Either an ace or not an ace.
3. The probability of success remains the same for each trial: The probability of drawing an ace remains constant throughout the trials.
4. The trials are independent: The outcome of one trial (drawing a card) does not affect the outcome of the next trial.
b. Probability of getting a pair of Aces:
To find the probability of getting a pair of Aces, we need to determine the probability of getting two aces out of the five cards drawn.
The probability of drawing the first ace is 4/52 since there are four aces in a deck of 52 cards.
After drawing the first ace, the probability of drawing the second ace is 3/51 since there are now only three aces left in the remaining 51 cards.
To calculate the probability of both events occurring, we multiply the individual probabilities together:
P(pair of Aces) = (4/52) * (3/51) = 1/221
c. Probability of getting exactly one Ace:
To find the probability of getting exactly one ace, we need to consider the different ways and arrangements in which one ace can be drawn.
There are five possible positions (first card, second card, third card, fourth card, or fifth card) where the ace can be drawn. We can calculate this probability by taking the sum of the probabilities of drawing an ace in each of the five possible positions.
The probability for each position is the same as the probability for the first card in the previous calculation: 4/52.
So, the probability of getting exactly one ace is:
P(exactly one Ace) = 5 * (4/52) * (48/51) * (47/50) * (46/49) * (45/48) ≈ 54.1%
d. Probability of getting no aces:
To find the probability of getting no aces in the five-card hand, we need to calculate the probability of not getting an ace for each of the five cards drawn.
The probability of not getting an ace on the first card is 48/52 since there are 48 non-ace cards remaining out of 52.
For each subsequent card, the probability of not getting an ace decreases since each successful non-ace card reduces the numerator by one.
To calculate the probability of no aces, we multiply the individual probabilities together for all five cards:
P(no Aces) = (48/52) * (47/51) * (46/50) * (45/49) * (44/48) ≈ 83.0%
e. Probability of getting at least one ace:
To find the probability of getting at least one ace, we can use the complement rule. The complement of getting at least one ace is getting no aces. Therefore, the probability of getting at least one ace is 1 minus the probability of getting no aces.
P(at least one Ace) = 1 - P(no Aces) = 1 - 0.83 ≈ 0.17, or 17%