Show that for all three distributions (Binomial, Poisson and Geometric) the probability of finding at least one ace is around 55%

To demonstrate that the probability of finding at least one ace is approximately 55% for all three distributions (Binomial, Poisson, and Geometric), we need to calculate the probabilities for each distribution individually.

1. Binomial Distribution:
The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials. In this case, the probability of finding at least one ace can be calculated using the binomial distribution with parameters n (number of trials) and p (probability of success).

Let's assume we have a standard deck of 52 playing cards, and we want to find the probability of drawing at least one ace if we draw 5 cards (n = 5). The probability of drawing an ace is 4/52, as there are four aces in the deck.

To calculate the probability of getting at least one ace, we can find the complement of the probability of getting zero aces (no aces at all) and subtract it from 1. So, the calculation is as follows:

P(at least one ace) = 1 - P(zero aces)
P(zero aces) = C(n, k) * p^k * (1 - p)^(n - k)

Using the binomial probability formula, we have:
P(at least one ace) = 1 - C(5, 0) * (4/52)^0 * (48/52)^5

Performing the calculations, we get:
P(at least one ace) = 1 - (1 * 1 * 0.447) = 0.553

Hence, the probability of finding at least one ace using the binomial distribution is approximately 55%.

2. Poisson Distribution:
The Poisson distribution is commonly used to model the number of events that occur in a fixed interval of time or space, given the average rate of occurrence. In this case, we can use the Poisson distribution to estimate the probability of finding at least one ace.

Assuming that the average number of aces in a 5-card hand is λ, we can calculate the probability of finding at least one ace using the Poisson distribution formula:

P(at least one ace) = 1 - P(zero aces)
P(zero aces) = e^(-λ) * (λ^k / k!)

To find the value of λ (average number of aces), we need to calculate the expected number of aces in a 5-card hand:
λ = expected number of aces per card * number of cards
= (4/52) * 5
= 0.385

Using the Poisson probability formula, we can calculate the probability of getting at least one ace:
P(at least one ace) = 1 - e^(-0.385) * (0.385^0 / 0!)
= 1 - 0.681
= 0.319

Therefore, the probability of finding at least one ace using the Poisson distribution is approximately 32%.

3. Geometric Distribution:
The geometric distribution models the number of trials required to achieve the first success in a sequence of independent Bernoulli trials. In this case, we can use the geometric distribution to calculate the probability of finding at least one ace.

Let's assume we draw cards one by one from a standard deck until we get the first ace. The probability of drawing an ace is 4/52, so the probability of not drawing an ace on the first draw is 48/52 (since we have 48 non-ace cards out of 52).

The probability of finding at least one ace can be calculated using the geometric probability formula:

P(at least one ace) = 1 - P(no ace before k trials)
P(no ace before k trials) = (1 - p)^(k - 1)

Using the geometric probability formula, we can calculate the probability of getting at least one ace:
P(at least one ace) = 1 - (48/52)^(k - 1)

We want to find the smallest value of k such that the probability of not getting an ace before k trials is less than 0.5 (since we are looking for a probability around 55%).

Let's calculate the probabilities for k = 1, 2, 3, 4, ... until we find the first k where P(no ace before k trials) is less than 0.5:

k = 1: P(no ace before 1 trial) = (48/52)^(1-1) = 1
k = 2: P(no ace before 2 trials) = (48/52)^(2-1) = 0.923
k = 3: P(no ace before 3 trials) = (48/52)^(3-1) = 0.894
k = 4: P(no ace before 4 trials) = (48/52)^(4-1) = 0.868
k = 5: P(no ace before 5 trials) = (48/52)^(5-1) = 0.844
k = 6: P(no ace before 6 trials) = (48/52)^(6-1) = 0.821
k = 7: P(no ace before 7 trials) = (48/52)^(7-1) = 0.801

We can observe that the probability of not getting an ace before the 7th trial is approximately 0.801. Therefore, the probability of finding at least one ace using the geometric distribution is approximately 1 - 0.801 = 0.199.

In conclusion, the probabilities of finding at least one ace for the Binomial, Poisson, and Geometric distributions are approximately 55%, 32%, and 20%, respectively.