Probability Of Drawing Cards A Step By Step Solution
In this engaging probability puzzle, we delve into the world of card draws and conditional probabilities. Hiro has a set of cards with numbers from the set {1, 1, 2, 2, 3, 3, 3, 4} written on them. Our task is to determine the probability that Hiro pulls out a 3 first, and then a 2, without replacing the cards. This problem allows us to explore the fundamental concepts of probability, conditional probability, and how to calculate probabilities in sequential events. Let's embark on this probabilistic journey and unravel the solution step by step.
Understanding the Problem
To effectively tackle this probability problem, we first need to grasp the core concepts involved. Probability, at its heart, is the measure of the likelihood of an event occurring. It's often expressed as a fraction, decimal, or percentage, representing the ratio of favorable outcomes to the total possible outcomes. In our case, the events are drawing a 3 first and then drawing a 2 from Hiro's set of cards.
Conditional probability comes into play when the occurrence of one event affects the probability of another event. This is precisely what happens when Hiro draws a card without replacing it. The first draw changes the composition of the remaining cards, thus influencing the probability of the second draw. Therefore, understanding conditional probability is crucial for accurately solving this problem.
Specifically, we have a set of cards: {1, 1, 2, 2, 3, 3, 3, 4}. We want to find the probability of two sequential events:
- Drawing a 3 first.
- Drawing a 2 second, without replacing the first card.
This means we're dealing with a situation where the outcome of the first event directly impacts the possible outcomes and probabilities of the second event. We will meticulously calculate the probability of each event, considering the changing state of the card set after each draw. This step-by-step approach ensures we accurately account for the conditional nature of the problem, leading us to the correct final probability.
Calculating the Probability of Drawing a 3 First
To determine the probability of drawing a 3 first, we need to consider the total number of cards and the number of cards with the number 3 on them. In Hiro's set of cards {1, 1, 2, 2, 3, 3, 3, 4}, there are a total of 8 cards. Among these, there are three cards with the number 3. Therefore, the probability of drawing a 3 on the first draw is the ratio of the number of 3s to the total number of cards.
This can be expressed as:
Probability (Drawing a 3 first) = (Number of cards with 3) / (Total number of cards)
Substituting the values, we get:
Probability (Drawing a 3 first) = 3 / 8
Thus, the probability that Hiro draws a 3 on his first attempt is 3/8. This fraction represents the likelihood of the first event occurring and serves as the foundation for calculating the subsequent conditional probability. Understanding this initial probability is crucial, as it directly influences the probability of drawing a 2 in the next step, given that a 3 has already been removed from the set. The next step will involve calculating the probability of drawing a 2, considering the reduced number of cards and the absence of one 3 from the deck.
Calculating the Probability of Drawing a 2 Second (Given a 3 Was Drawn First)
Now, we move on to the crucial step of calculating the probability of drawing a 2 on the second draw, given that a 3 was drawn first and not replaced. This is where the concept of conditional probability comes into play. After Hiro draws a 3, the total number of cards in the set is reduced by one, and the number of 3s is also reduced by one. This changes the composition of the remaining cards and thus affects the probabilities for the next draw.
Initially, there were 8 cards in total. After drawing a 3, there are now only 7 cards remaining. The set of cards now looks like this (assuming one 3 has been removed): {1, 1, 2, 2, 3, 3, 4}. We are interested in the probability of drawing a 2 from this reduced set.
In the original set, there were two cards with the number 2. Since no 2s were removed in the first draw, there are still two 2s present in the remaining set of 7 cards. Therefore, the probability of drawing a 2 on the second draw, given that a 3 was drawn first, is the ratio of the number of 2s to the new total number of cards.
This can be expressed as:
Probability (Drawing a 2 second | Drawing a 3 first) = (Number of 2s remaining) / (Total number of cards remaining)
Substituting the values, we get:
Probability (Drawing a 2 second | Drawing a 3 first) = 2 / 7
This fraction, 2/7, represents the conditional probability of drawing a 2 after a 3 has already been drawn. It reflects the altered state of the card set and the impact of the first event on the likelihood of the second. In the next step, we will combine this conditional probability with the initial probability of drawing a 3 to find the overall probability of both events occurring in sequence.
Combining the Probabilities
To find the overall probability of Hiro drawing a 3 first and then a 2 without replacement, we need to combine the probabilities of the individual events. Since these events are sequential and the second event's probability is conditional on the first, we use the rule of multiplying probabilities for dependent events.
The rule states that the probability of two dependent events A and B occurring is:
Probability (A and B) = Probability (A) * Probability (B | A)
In our case:
- Event A is drawing a 3 first.
- Event B is drawing a 2 second.
We have already calculated:
- Probability (Drawing a 3 first) = 3 / 8
- Probability (Drawing a 2 second | Drawing a 3 first) = 2 / 7
Now, we multiply these probabilities together to find the overall probability:
Probability (Drawing a 3 first and then a 2) = (3 / 8) * (2 / 7)
Multiplying the numerators and denominators, we get:
Probability (Drawing a 3 first and then a 2) = 6 / 56
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2:
Probability (Drawing a 3 first and then a 2) = 3 / 28
Therefore, the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacing them is 3/28. This result is the culmination of our step-by-step analysis, accurately accounting for the conditional nature of the problem and providing the final solution to the probability puzzle.
Final Answer
After carefully calculating the probabilities of each event and combining them appropriately, we have arrived at the solution. The probability that Hiro pulls out a 3 first and then pulls out a 2 without replacing the cards is:
3/28
This answer aligns with one of the provided options, confirming the accuracy of our step-by-step approach. By breaking down the problem into manageable parts and meticulously applying the principles of probability and conditional probability, we have successfully navigated this card-drawing puzzle. The process has highlighted the importance of understanding conditional probabilities when dealing with sequential events, where the outcome of one event directly influences the probabilities of subsequent events.
