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Chapter 3: Problem 10

Three cards are randomly selected, without replacement, from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade, given that the second and third cards are spades.

Short Answer

Expert verified
The conditional probability that the first card selected is a spade, given that the second and third cards are spades, is \(\frac{13}{52}\) or approximately 0.25 (25%).

Step by step solution

01

Find the probability of the first, second, and third cards being spades

Since there are 13 spades in the deck of 52 cards, the probability that the first card is a spade is 13/52. Given that the first card is a spade, there are now 12 spades left in the deck of 51 cards, so the probability that the second card is a spade is 12/51. Similarly, the probability that the third card is a spade, given that the first two cards were spades, is 11/50. The probability of all three events happening (first, second, and third cards being spades) is the product of these individual probabilities: P(first, second, and third cards are spades) = \(\frac{13}{52} \times \frac{12}{51} \times \frac{11}{50}\)
02

Find the probability of the second and third cards being spades

The first card can be any of the remaining 51 cards, and we have 12 remaining spades in a 51-card deck. The probability that the second card is a spade is then 12/51. Given that the second card is a spade, there are 11 spades left in a 50-card deck, so the probability that the third card is a spade is 11/50. The probability that the second and third cards are spades (ignoring the condition on the first card) is the product of their individual probabilities: P(second and third cards are spades) = \(\frac{12}{51} \times \frac{11}{50}\)
03

Calculate the conditional probability

Now, we can use the conditional probability formula: P(A|B) = P(A and B) / P(B) P(first card is a spade | second and third cards are spades) = \(\frac{P(\text{first, second, and third cards are spades})}{P(\text{second and third cards are spades})}\) = \(\frac{\frac{13}{52} \times \frac{12}{51} \times \frac{11}{50}}{\frac{12}{51} \times \frac{11}{50}}\) By canceling out common terms, we get: P(first card is a spade | second and third cards are spades) = \(\frac{13}{52}\) So, the conditional probability that the first card selected is a spade, given that the second and third cards are spades, is 13/52 or approximately 0.25 (25%).

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of events occurring. It provides a mathematical framework to quantify uncertainty and helps us make informed decisions based on incomplete information. In the context of the exercise, probability theory is applied to a deck of playing cards, a common example in the study of probability.

Why is this important? When we calculate probabilities, we consider all possible outcomes and evaluate how likely a particular event is within this set. The deck of playing cards has a fixed number of possible outcomes, making it an ideal model for learning basic probability concepts. In our example, the event of interest is drawing a spade from a standard 52-card deck.
Combinatorics
Combinatorics is the field of mathematics focused on counting, arranging, and finding patterns within sets of elements. It becomes crucial in probability theory when we need to establish the number of ways certain events can occur. For example, figuring out the number of possible 3-card combinations from a deck is a combinatorial problem.

In our exercise, combinatorics come into play when considering the various combinations of cards drawn and the decreasing number of cards as each card is drawn without replacement. It helps us understand the relationship between the remaining cards in the deck and the changing probabilities with each draw. Mastering combinatorics is key to accurately calculating compound probabilities in more complex scenarios.
Probability Without Replacement
When we talk about 'probability without replacement', we refer to the scenario where each item is not returned to the set after it is picked, therefore changing the probability of subsequent events. This concept is different from 'probability with replacement', where each item is returned to the set, keeping the probabilities consistent across trials.

Our example specifically deals with the probability without replacement. As cards are drawn from the deck, the total number of cards and the number of spades decrease, altering the probabilities of future draws. The solution to the exercise cleverly illustrates this concept by showing how the probability of drawing a spade changes from one draw to the next, and how it affects the calculation of the conditional probability.

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Most popular questions from this chapter

Independent trials that result in a success with probability \(p\) are successively performed until a total of \(r\) successes is obtained. Show that the probability that exactly \(n\) trials are required is $$ \left(\begin{array}{l} n-1 \\ r-1 \end{array}\right) p^{r}(1-p)^{n-r} $$ Use this result to solve the problem of the points (Example 4i). HINT. In order for it to take \(n\) trials to obtain \(r\) successes, how many successes must occur in the first \(n-1\) trials?

Fifty-two percent of the students at a certain college are females. Five percent of the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that (a) this student is female, given that the student is majoring in computer science; (b) this student is majoring in computer science, given that the student is female.

A ball is in any one of \(n\) boxes. It is in the \(i\) th box with probability \(P_{i}\). If the ball is in box \(i\), a search of that box will uncover it with probability \(\alpha_{i}\). Show that the conditional probability that the ball is in box \(j\), given that a search of box \(i\) did not uncover it, is $$ \begin{array}{cl} \frac{P_{j}}{1-\alpha_{i} P_{i}} & \text { if } j \neq i \\ \frac{\left(1-\alpha_{i}\right) P_{i}}{1-\alpha_{i} P_{i}} & \text { if } j=i \end{array} $$

There are 3 coins in a box. One is a two-headed coin; another is a fair coin; and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

Consider the gambler's ruin problem with the exception that \(A\) and \(B\) agree to play no more than \(n\) games. Let \(P_{n, i}\) denote the probability that \(A\) winds up with all the money when \(A\) starts with \(i\) and \(B\) with \(N-i\). Derive an equation for \(P_{n, i}\) in terms of \(P_{n-1, i+1}\) and \(P_{n-1, i-1}\) and compute \(P_{7,3}\), \(N=5\)
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