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Chapter 4: Problem 23

One card is drawn from each of two ordinary packs of 52 cards. The probability that at least one of them is an ace of heart is (1) \(\frac{103}{2704}\) (2) \(\frac{1}{2704}\) (3) \(\frac{2}{52}\) (4) \(\frac{2601}{2704}\) (5) \(\frac{51}{52}\)

Short Answer

Expert verified
The probability that at least one of the cards drawn is an ace of hearts is \( \frac{103}{2704} \).

Step by step solution

01

- Identify the Total Possible Outcomes

Each deck has 52 cards, so the total number of possible outcomes when drawing one card from each of the two decks is: Total outcomes = 52 * 52 = 2704
02

- Calculate the Probability of Not Drawing an Ace of Heart

First, find the probability of not drawing an ace of heart from each deck. There are 51 cards that are not an ace of heart in each deck.Probability of not drawing an ace of heart from one deck = \( \frac{51}{52} \)For two decks, the combined probability of not drawing an ace of heart from both is:\( \left( \frac{51}{52} \right) \times \left( \frac{51}{52} \right) = \frac{2601}{2704} \)
03

- Calculate the Probability of Drawing at Least One Ace of Heart

To find the probability of drawing at least one ace of heart, subtract the probability of not drawing an ace of heart from 1:Probability = 1 - \( \frac{2601}{2704} \)= \( \frac{2704}{2704} - \frac{2601}{2704} \)= \( \frac{103}{2704} \)

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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 all about measuring the likelihood of events occurring.
This likelihood is something we represent with numbers between 0 and 1.
A probability of 0 means the event will not happen, while a probability of 1 means it definitely will. The core formula for probability is:
\( \text{Probability of an event} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
In the exercise problem, we calculate multiple probabilities. Starting with the total possible outcomes when drawing one card from each of the two decks, and then moving to the probability of not drawing an ace of heart.
combinatorics
Combinatorics deals with counting and arranging objects.
This helps us understand all possible outcomes in a situation.
For our exercise, we deal with the total number of outcomes when drawing cards.
Each deck has 52 cards, so the total outcomes when drawing one card from each of the two decks is:
\( 52 \times 52 = 2704 \)
From here, we figure out the number of ways not to get an ace of hearts. Each deck has 51 cards that are not the ace of hearts.
Therefore, the combined probability of not drawing an ace of heart from either deck is:
\( \left( \frac{51}{52} \right) \times \left( \frac{51}{52} \right) = \frac{2601}{2704} \)
ace of hearts probability
To determine if we draw at least one ace of hearts, we use the complementary probability concept.
This involves subtracting the probability of not getting an ace of hearts from 1. Since:
\( \text{Probability} = 1 - \frac{2601}{2704} \)
Therefore, the probability of drawing at least one ace of hearts is:
\(1 - \frac{2601}{2704} = \frac{2704}{2704} - \frac{2601}{2704} = \frac{103}{2704}\).

So, if you're trying to find the likelihood of drawing at least one ace of hearts from two decks, you now know it's \( \frac{103}{2704} \)

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Most popular questions from this chapter

STATEMENT-1 : A card is drawn from a well shuffled ordinary deck of 52 -playing cards. Let \(\mathrm{A}\) be the event that 'card drawn is an Ace' and \(\mathrm{B}\) be the event that 'card drawn is a spade'. Then the events \(A\) and \(B\) are indpendent STATEMENT-2 : Let \(A\) and \(B\) be two non-empty events. If \(P(A / B)=P(A)\), then the events \(A\) and \(B\) are indpendent. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (3) Statement- 1 is True, Statement-2 is False (4) Statement- 1 is False, Statement- 2 is True

The total number of ways of selecting 10 balls out of an unlimited number of identical white, red and blue balls is equal to (1) \({ }^{12} \mathrm{C}_{2}\) (2) \({ }^{13} \mathrm{C}_{3}\) (3) \({ }^{10} \mathrm{C}_{2}\) (4) \({ }^{10} \mathrm{C}_{3}\) (5) \({ }^{12} \mathrm{C}_{3}\)

A river is flowing with a speed of \(1 \mathrm{~km} / \mathrm{hr}\). \(\mathrm{A}\) swimmer wants to go to point 'C' starting from 'A'. He swims with a speed of \(5 \mathrm{~km} / \mathrm{hr}\), at an angle \(\theta\) w.r.t. the river. If \(A B=B C=400 \mathrm{~m}\). Then the value of \(\theta\) is: (1) \(37^{\circ}\) (2) \(30^{\circ}\) (3) \(60^{\circ}\) (4) \(53^{\circ}\)

Two concentric rings, one of radius \(\mathrm{R}\) and total charge \(+\mathrm{Q}\) and the second of radius \(2 \mathrm{R}\) and total charge \(-\sqrt{8} \mathrm{Q}\), lie in \(\mathrm{x}-\mathrm{y}\) plane (i.e., \(\mathrm{z}=0\) plane). The common centre of rings lies at origin and the common axis coincides with z-axis. The charge is uniformly distributed on both rings. The net electric field on z-axis is zero at a distance \(\frac{8 \mathrm{R}}{(\sqrt{2})^{x}}\) from origin. Then find \(\mathrm{x}\). (1) 2 (2) 4 (3) 6 (4) 8 (5) 5

A solution of \(0.2\) mole \(\mathrm{KI}(\alpha=100 \%)\) in 1000 \(\mathrm{g}\) water freezes at \(\mathrm{T}_{1}{ }^{\circ} \mathrm{C}\). Now to this solution \(0.1\) mole \(\mathrm{HgI}_{2}\) is added and the resulting solution freezes at \(\mathrm{T}_{2}{ }^{\circ} \mathrm{C}\). Which of the following is correct: (1) \(T_{1}=T_{2}\) (2) \(\mathrm{T}_{1}>\mathrm{T}_{2}\) (3) \(\mathrm{T}_{1}<\mathrm{T}_{2}\) (4) Can't be prediced
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