Probability of Drawing Specific Cards with Replacement vs. Without Replacement

Consider a scenario where you have 10 cards labeled 1 through 10. You draw 4 cards with replacement. What is the probability that two of the drawn cards are cards labeled 2 and 4? This problem can be approached both with and without replacement, leading to different probabilities. Let's explore both scenarios in detail.

Scenario with Replacement

Possible Outcomes:

When drawing cards with replacement, each card has an equal probability of being drawn in each of the four draws. The total number of possible outcomes is calculated as follows:

For each of the 4 draws, there are 10 possible choices. Thus, the total number of outcomes is (10^4 10000).

Favorable Outcomes:

To have exactly two out of the four drawn cards be cards 2 and 4, we need to consider the different ways in which these two cards can appear among the four draws. The number of favorable sequences is calculated by considering the placement of the cards 2 and 4 among the four draws, and then the remaining two draws can be any of the 10 cards.

The number of ways to choose 2 draws out of 4 for cards 2 and 4 is given by the binomial coefficient ( binom{4}{2} ), which equals 6.

For each of these 6 combinations, the remaining 2 draws can be any of the 10 cards, giving (10^2 100) possibilities. Thus, the total number of favorable outcomes is (6 times 100 600).

Probability Calculation:

The probability of this event is the ratio of the number of favorable outcomes to the total number of possible outcomes:

[ P(text{2 and 4 exactly twice}) frac{600}{10000} 0.06 ]

Scenario Without Replacement

Exact Probabilities Using Hypergeometric Distribution:

Given that drawing without replacement means we do not put the card back after each draw, the situation is more complex. We can use the Hypergeometric Distribution to calculate the exact probability.

In this scenario:

Population Size: 10 cards Successes in Population: 2 cards labeled 2 and 4 Sample Size: 4 cards drawn Successes in Sample: Exactly 2 cards labeled 2 and 4

The Hypergeometric Distribution formula for this specific scenario is:

[ P(Xk) frac{binom{K}{k} binom{N-K}{n-k}}{binom{N}{n}} ]

Where:

N: Population size (10) K: Number of success states in the population (2) n: Number of draws (4) k: Number of observed successes (2)

Plugging these values into the formula, we get:

[ P(X2) frac{binom{2}{2} binom{10-2}{4-2}}{binom{10}{4}} frac{binom{2}{2} binom{8}{2}}{binom{10}{4}} ]

[ P(X2) frac{1 times 28}{210} frac{28}{210} frac{4}{30} frac{2}{15} approx 0.1333 ]

Conclusion:

The probability of drawing exactly two cards labeled 2 and 4, given the scenario of drawing with replacement is 0.06, whereas the exact probability under the scenario of drawing without replacement is approximately 0.1333. The difference arises from the fact that drawing without replacement changes the probabilities over the course of the draws.

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