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Cards are dealt, one at a time, from a standard 52 -card deck. a. If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades? b. If the first 3 cards are all spades, what is the probability that the next 2 cards are also spades? c. If the first 4 cards are all spades, what is the probability that the next card is also a spade?

Short Answer

Expert verified
a. \( \frac{11}{1300} \), b. \( \frac{15}{392} \), c. \( \frac{3}{16} \).

Step by step solution

01

Understand the problem

We are dealing with the problem of calculating probabilities involving dependent events since cards are removed without replacement. Each part of the question corresponds to a sub-problem of calculating a specific conditional probability.
02

Calculate Probability for (a)

First, we know the first 2 cards are spades, so 2 spades have already been dealt. There are 11 spades left and 50 cards remaining. The probability of the next 3 cards being spades:First spade: \( \frac{11}{50} \) Second spade: \( \frac{10}{49} \) Third spade: \( \frac{9}{48} \)Multiply these probabilities to find the total probability: \[ P(A) = \frac{11}{50} \times \frac{10}{49} \times \frac{9}{48} = \frac{990}{117600} = \frac{11}{1300} \approx 0.00846 \]
03

Calculate Probability for (b)

For the first 3 cards being spades, there are now 10 spades left in 49 cards. We need 2 more spades:First spade: \( \frac{10}{49} \)Second spade: \( \frac{9}{48} \)Multiply these probabilities to find the total probability:\[ P(B) = \frac{10}{49} \times \frac{9}{48} = \frac{90}{2352} = \frac{15}{392} \approx 0.03827 \]
04

Calculate Probability for (c)

Given the first 4 cards are spades, there are 9 spades left in the deck of 48 cards. We find the probability of the next card being a spade:Probability: \( \frac{9}{48} = \frac{3}{16} = 0.1875 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
In probability theory, conditional probability refers to the likelihood of an event occurring given that another event has already happened. This concept is essential when dealing with problems that involve sequential events where past outcomes affect future probabilities.

To calculate conditional probability, we use the formula:
  • \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \)
Here, \( P(A \mid B) \) represents the probability of event A occurring given that B has occurred. \( P(A \cap B) \) is the probability that both events happen, and \( P(B) \) is the probability of event B.

In the context of the card problem, knowing that a certain number of spades have already been drawn affects the probability of drawing more spades. Hence, each task in the exercise is about finding the conditional probability of drawing additional spades based on previous draws.
Dependent Events
Dependent events are those where the outcome or occurrence of the first event affects the outcome or occurrence of the second event. When dealing cards from a deck without replacement, the probability of future events changes with each card dealt.

For example, if you've drawn one spade from a deck, the probability of drawing another spade changes because there is now one fewer spade in the deck and one fewer card in total. This dependency requires careful adjustment of probabilities in sequential scenarios. When calculating probabilities involving dependent events, we use the modified number of outcomes and favorable cases.

In our card problem, after each spade is drawn, the number of available spades decreases, which demonstrates how the events are dependent on previous occurrences. As shown in the solution, this is why we multiply probabilities of each subsequent event with adjusted counts of total and favorable outcomes.
Combinatorics
Combinatorics is a field of mathematics that studies counting, combination, and permutation of sets. It plays a crucial role in calculating probabilities, especially when we deal with large sets like a deck of cards.

In the card example, combinatorics comes into play when determining how many ways the cards can be drawn. Calculating the likelihood of multiple dependent events—such as drawing sequential spades—relies heavily on combinatorial principles.

When we multiply probabilities of subsequent draws, we essentially use combinations to find the probability of drawing a specific sequence from the deck. Understanding how to count and arrange subsets of a larger set is vital in determining the overall probability, as demonstrated by sequentially reducing the number of spades and cards available.

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