Problem 7
Question
A person draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues doing this until be draws a spade, the chance that he will fail in the first two draws is (A) \(\frac{1}{16}\) (B) \(\frac{9}{16}\) (C) \(\frac{9}{64}\) (D) \(\frac{1}{64}\)
Step-by-Step Solution
Verified Answer
The chance he will fail in the first two draws is \( \frac{9}{16} \) (Option B).
1Step 1: Understanding the Probability of Drawing a Spade
A standard deck of playing cards has 52 cards, consisting of 4 suits: spades, hearts, diamonds, and clubs. Each suit has 13 cards. Therefore, the probability of drawing a spade in a single draw is calculated by dividing the number of spades (13) by the total number of cards (52). This gives us a probability of \( \frac{13}{52} = \frac{1}{4} \).
2Step 2: Probability of Failing to Draw a Spade
Failing to draw a spade means drawing any of the other suits (hearts, diamonds, or clubs). There are 39 cards that are not spades. Therefore, the probability of not drawing a spade in a single draw is \( \frac{39}{52} = \frac{3}{4} \).
3Step 3: Calculating Failure in the First Two Draws
Since the draw is with replacement, each draw is independent. To find the probability of failing to draw a spade in the first two draws, we multiply the probability of failing in one draw by itself: \( \left( \frac{3}{4} \right)^2 \).
4Step 4: Compute the Final Probability
Calculate \( \left( \frac{3}{4} \right)^2 \): \[ \left( \frac{3}{4} \right)^2 = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \].
Key Concepts
Probability with ReplacementDependent Events ProbabilityBasic Probability Concepts
Probability with Replacement
When dealing with the concept of probability, it's common to encounter problems involving drawing objects, like cards, from a set, and then replacing them. This is known as "Probability with Replacement."
In the original exercise, after drawing a card from a deck, the card is placed back into the deck and the deck is shuffled again. This replacement is crucial because it keeps the probabilities of each draw independent of each previous one. This means the total number of cards remains the same — 52 cards after each draw.
This replacement condition makes the probability of drawing any particular card stay consistent with each draw. For example, the probability of drawing a spade is always \(\frac{1}{4}\) for every draw, as there are always 13 spades out of 52 cards in the deck. Similarly, the probability of failing to draw a spade — drawing any other suit — is \(\frac{3}{4}\) with each draw. This constant probability is due to the replacement of the cards.
In the original exercise, after drawing a card from a deck, the card is placed back into the deck and the deck is shuffled again. This replacement is crucial because it keeps the probabilities of each draw independent of each previous one. This means the total number of cards remains the same — 52 cards after each draw.
This replacement condition makes the probability of drawing any particular card stay consistent with each draw. For example, the probability of drawing a spade is always \(\frac{1}{4}\) for every draw, as there are always 13 spades out of 52 cards in the deck. Similarly, the probability of failing to draw a spade — drawing any other suit — is \(\frac{3}{4}\) with each draw. This constant probability is due to the replacement of the cards.
Dependent Events Probability
In probability, events are considered dependent if the outcome of one event affects the outcome of another. However, in this exercise, because of the "with replacement" condition, each draw is independent, which is key to how we calculate the probability of events.
But to draw a parallel, let's consider what happens if events were dependent. Imagine drawing a card from the deck and not replacing it before drawing again. In this scenario, after drawing a card, the total number of cards changes — reducing to 51 in the second draw. This affects the subsequent probabilities because the deck composition changes.
For instance, if you first draw a spade and do not replace it, the probability of drawing another spade changes because there are now only 12 spades and 51 total cards left. So in problems without replacement, the probability calculations must account for this change in the composition of the set.
But to draw a parallel, let's consider what happens if events were dependent. Imagine drawing a card from the deck and not replacing it before drawing again. In this scenario, after drawing a card, the total number of cards changes — reducing to 51 in the second draw. This affects the subsequent probabilities because the deck composition changes.
For instance, if you first draw a spade and do not replace it, the probability of drawing another spade changes because there are now only 12 spades and 51 total cards left. So in problems without replacement, the probability calculations must account for this change in the composition of the set.
Basic Probability Concepts
Understanding basic probability concepts is crucial for solving exercises like these. Probability, in its essence, is the measure of how likely an event is to occur. It can be expressed as a fraction between 0 and 1, where 0 indicates impossibility and 1 certainty.
The probability of an event \( A \) is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes. So, when we find the probability of drawing a spade from a deck of cards, it is computed as \( \frac{13}{52} \) since there are 13 favorable outcomes (spades) out of 52 possible outcomes (total cards).
Another fundamental rule is that the probability of all possible outcomes in a sample space adds up to 1. This means the probability of drawing a spade (\( \frac{1}{4} \)) and not drawing a spade (\( \frac{3}{4} \)) must equal 1. These concepts allow us to solve more complex probability problems, like determining the chance of multiple consecutive events, such as drawing multiple non-spades in a row, by multiplying the probabilities of individual independent events.
The probability of an event \( A \) is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes. So, when we find the probability of drawing a spade from a deck of cards, it is computed as \( \frac{13}{52} \) since there are 13 favorable outcomes (spades) out of 52 possible outcomes (total cards).
Another fundamental rule is that the probability of all possible outcomes in a sample space adds up to 1. This means the probability of drawing a spade (\( \frac{1}{4} \)) and not drawing a spade (\( \frac{3}{4} \)) must equal 1. These concepts allow us to solve more complex probability problems, like determining the chance of multiple consecutive events, such as drawing multiple non-spades in a row, by multiplying the probabilities of individual independent events.
Other exercises in this chapter
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