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Chapter 7: Problem 7

Two cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. Find the probability of the given event. Two black cards are drawn.

Short Answer

Expert verified
The probability of drawing two black cards without replacement from a well-shuffled deck of 52 playing cards is \(\frac{325}{2652}\).

Step by step solution

01

Calculate the probability of drawing the first black card

There are 26 black cards in a deck of 52 cards. Therefore, the probability of drawing the first black card is the ratio of the number of black cards to the total number of cards in the deck: \(P(\text{first black card}) = \frac{26}{52}\)
02

Calculate the probability of drawing the second black card

Since we have drawn one black card and we're not replacing it, there are now 25 black cards left in the deck and the deck has a total of 51 cards. Therefore, the probability of drawing the second black card is the ratio of the number of remaining black cards to the total number of cards in the deck minus one: \(P(\text{second black card}|\text{first black card}) = \frac{25}{51}\)
03

Multiply the probabilities together

Now, we need to multiply the probability of drawing the first black card and the probability of drawing the second black card to get the final probability: \(P(\text{two black cards}) = P(\text{first black card}) \times P(\text{second black card}|\text{first black card})\) \(P(\text{two black cards}) = \frac{26}{52} \times \frac{25}{51}\)
04

Simplify the expression and find the probability

Simplify the fraction to find the probability: \(P(\text{two black cards}) = \frac{13}{52} \times \frac{25}{51}\) \(P(\text{two black cards}) = \frac{13 \times 25}{52 \times 51}\) \(P(\text{two black cards}) = \frac{325}{2652}\) So, the probability of drawing two black cards without replacement is \(\frac{325}{2652}\).

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arranging, and grouping objects. It helps in identifying how many ways a certain event can occur.
For example, when drawing cards from a deck, combinatorics aids in calculating the total possible outcomes of drawing two cards consecutively without replacement.
This calculation is essential in determining the number of favorable outcomes for drawing two black cards from a deck of 52 cards:
  • There are 26 black cards in a deck.
  • When one is drawn, only 25 are left for the next draw.
This is where understanding combinations and permutations becomes crucial. In our problem, we used these concepts implicitly by counting favorable outcomes without explicitly rearranging or listing them.
Conditional Probability
Conditional probability deals with the likelihood of an event occurring given that another event has already happened. It becomes especially important when dealing with events that are impacted by prior actions鈥攕uch as in our exercise of drawing cards without replacement.
When the first black card is drawn, the probability of the second card being black changes:
  • The probability of the first black card: \( \frac{26}{52} \)
  • Once a black card is drawn, 25 black cards remain, changing the sample space to 51 cards: \( \frac{25}{51} \).
To find the probability of both events occurring (drawing two black cards in succession), these probabilities are multiplied. This multiplication reflects the dependent nature of the second event on the first.
Deck of Cards
A standard deck of cards is a common tool in probability exercises due to its fixed number of outcomes and the uniformity of its distribution.
The deck consists of 52 cards, divided evenly into four suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards.
This structure is key to understanding probabilities related to drawing cards:
  • There are 26 red cards (hearts and diamonds).
  • There are 26 black cards (clubs and spades).
Knowing this structure helps in calculating the probability of drawing certain types of cards, like in the given exercise where we focus on black cards. Recognizing the total number and breakdown of the deck ensures precise probability calculations by maintaining correct counts of available options after each draw.

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Most popular questions from this chapter

In an online survey of 1962 executives from 64 countries conducted by Korn/Ferry International between August and October 2006, the executives were asked if they would try to influence their children's career choices. Their replies: A (to a very great extent), B (to a great extent), \(\mathrm{C}\) (to some extent), D (to a small extent), and \(\mathrm{E}\) (not at all) are recorded below: $$\begin{array}{lccccc} \hline \text { Answer } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\ \hline \text { Respondents } & 135 & 404 & 1057 & 211 & 155 \\ \hline \end{array}$$ What is the probability that a randomly selected respondent's answer was \(\mathrm{D}\) (to a small extent) or \(\mathrm{E}\) (not at all)?

Determine whether the given experiment has a sample space with equally likely outcomes. A loaded die is rolled, and the number appearing uppermost on the die is recorded.

The personnel department of Franklin National Life Insurance Company compiled the accompanying data regarding the income and education of its employees: \begin{tabular}{lcc} \hline & Income & Income \\ & \(\$ 50,000\) or Relow & Above \$50,000 \\ \hline Noncollege Graduate & 2040 & 840 \\ \hline College Graduate & 400 & 720 \\ \hline \end{tabular} Let \(A\) be the cvent that a randomly chosen cmployee has a college degree and \(B\) the cvent that the chosen cmployec's income is more than \(\$ 50.000\). a. Find cach of the following probabilities: \(P(A), P(B)\), \(P(A \cap B), P(B \mid A)\), and \(P\left(B \mid A^{c}\right)\) b. Are the cvents \(A\) and \(B\) independent events?

In a survey of 2000 adults \(50 \mathrm{yr}\) and older of whom \(60 \%\) were retired and \(40 \%\) were preretired, the following question was asked: Do you expect your income needs to vary from year to year in retirement? Of those who were retired, \(33 \%\) answered no, and \(67 \%\) answered yes. Of those who were pre-retired, \(28 \%\) answered no, and \(72 \%\) answered yes. If a respondent in the survey was selected at random and had answered yes to the question, what is the probability that he or she was retired?

Asurvey involving 400 likely Democratic voters and 300 likely Republican voters asked the question: Do you support or oppose legislation that would require registration of all handguns? The following results were obtained: $$\begin{array}{lcc} \hline \text { Answer } & \text { Democrats, \% } & \text { Republicans, \% } \\\ \hline \text { Support } & 77 & 59 \\ \hline \text { Oppose } & 14 & 31 \\ \hline \text { Don't know/refused } & 9 & 10 \\ \hline \end{array}$$ If a randomly chosen respondent in the survey answered "oppose," what is the probability that he or she is a likely Democratic voter?
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