Question
There are 100 trading cards in a bin. They include 5 rare cards, 35 limited cards, and the remainder are common cards. Which of the following correctly interprets the probability of choosing 2 rare cards from the bin? Assume that the cards are pulled without replacement.
A.
100
5
⋅
100
5
B.
100
5
⋅
99
4
C.
100
5
+
100
5
D.
100
5
A.
100
5
⋅
100
5
B.
100
5
⋅
99
4
C.
100
5
+
100
5
D.
100
5
Answers
GPT-4o mini
To find the probability of choosing 2 rare cards from a set of 100 trading cards, without replacement, we can break down the selection process as follows:
1. **Choose the first rare card**: The probability of choosing a rare card first is the number of rare cards divided by the total number of cards, which is \(\frac{5}{100}\).
2. **Choose the second rare card**: After choosing the first rare card, there will be only 4 rare cards left and 99 total cards remaining. Therefore, the probability of choosing a second rare card is \(\frac{4}{99}\).
To get the probability of both events happening (choosing two rare cards in a row), we multiply these probabilities together:
\[
\text{Probability} = \frac{5}{100} \times \frac{4}{99}
\]
This expression simplifies to:
\[
\text{Probability} = \frac{5 \cdot 4}{100 \cdot 99} = \frac{20}{9900} = \frac{1}{495}
\]
Now, let's match this conclusion with the provided options.
We should express it in terms similar to those options:
- The number of ways to choose 2 rare cards from 5 for the first and second selection would be represented in terms of combinations, not fractions.
Looking closely, we see that:
- Option A: \(\frac{100}{5} \cdot \frac{100}{5}\)
- Option B: \(\frac{100}{5} \cdot \frac{99}{4}\)
- C: \(\frac{100}{5} + \frac{100}{5}\)
- D: \(\frac{100}{5}\)
Since the probabilities of choosing two rare cards in that specific order without replacement involves multiplying the probabilities of individual selections (as noted above), the best choice that follows the same logic of selecting two items from the scenario above is **Option B**:
\[
\frac{100}{5} \cdot \frac{99}{4}
\]
So the correct interpretation of the probability of choosing 2 rare cards from the bin is:
**B. \( \frac{100}{5} \cdot \frac{99}{4} \)**
1. **Choose the first rare card**: The probability of choosing a rare card first is the number of rare cards divided by the total number of cards, which is \(\frac{5}{100}\).
2. **Choose the second rare card**: After choosing the first rare card, there will be only 4 rare cards left and 99 total cards remaining. Therefore, the probability of choosing a second rare card is \(\frac{4}{99}\).
To get the probability of both events happening (choosing two rare cards in a row), we multiply these probabilities together:
\[
\text{Probability} = \frac{5}{100} \times \frac{4}{99}
\]
This expression simplifies to:
\[
\text{Probability} = \frac{5 \cdot 4}{100 \cdot 99} = \frac{20}{9900} = \frac{1}{495}
\]
Now, let's match this conclusion with the provided options.
We should express it in terms similar to those options:
- The number of ways to choose 2 rare cards from 5 for the first and second selection would be represented in terms of combinations, not fractions.
Looking closely, we see that:
- Option A: \(\frac{100}{5} \cdot \frac{100}{5}\)
- Option B: \(\frac{100}{5} \cdot \frac{99}{4}\)
- C: \(\frac{100}{5} + \frac{100}{5}\)
- D: \(\frac{100}{5}\)
Since the probabilities of choosing two rare cards in that specific order without replacement involves multiplying the probabilities of individual selections (as noted above), the best choice that follows the same logic of selecting two items from the scenario above is **Option B**:
\[
\frac{100}{5} \cdot \frac{99}{4}
\]
So the correct interpretation of the probability of choosing 2 rare cards from the bin is:
**B. \( \frac{100}{5} \cdot \frac{99}{4} \)**