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.(1 point) Responses 5100⋅5100 Start Fraction 5 over 100 End Fraction times Start Fraction 5 over 100 End Fraction 5100+5100 Start Fraction 5 over 100 End Fraction times Start Fraction 5 over 100 End Fraction 5100 Start Fraction 5 over 100 End Fraction 5100⋅499 Start Fraction 5 over 100 End Fraction times Start Fraction 4 over 99 End Fraction
Answers
GPT-4o mini
To calculate the probability of choosing 2 rare cards from the bin without replacement, we can break this down into two parts:
1. The probability of picking the first rare card.
2. The probability of picking the second rare card, given that the first one was already picked.
1. The probability of picking the first rare card:
There are 5 rare cards out of a total of 100 cards, so the probability is:
\[
\frac{5}{100}
\]
2. The probability of picking the second rare card after one rare card has already been picked:
After picking the first rare card, there are now 4 rare cards left and only 99 cards remaining total, so the probability is:
\[
\frac{4}{99}
\]
Now, to find the overall probability of both events happening (i.e., picking two rare cards in sequence without replacement), we multiply the two probabilities:
\[
\text{Probability of choosing 2 rare cards} = \frac{5}{100} \times \frac{4}{99}
\]
Therefore, the correct response from the options given would be:
\[
5100 \cdot 499 \frac{5}{100} \times \frac{4}{99}
\]
However, since "5100" and "5100⋅499" are not standard notation for expressing the probabilities and don't quite fit here directly, the clearest answer to select based on the interpretation of choosing two rare cards without replacement is:
\[
\frac{5}{100} \times \frac{4}{99}
\]
1. The probability of picking the first rare card.
2. The probability of picking the second rare card, given that the first one was already picked.
1. The probability of picking the first rare card:
There are 5 rare cards out of a total of 100 cards, so the probability is:
\[
\frac{5}{100}
\]
2. The probability of picking the second rare card after one rare card has already been picked:
After picking the first rare card, there are now 4 rare cards left and only 99 cards remaining total, so the probability is:
\[
\frac{4}{99}
\]
Now, to find the overall probability of both events happening (i.e., picking two rare cards in sequence without replacement), we multiply the two probabilities:
\[
\text{Probability of choosing 2 rare cards} = \frac{5}{100} \times \frac{4}{99}
\]
Therefore, the correct response from the options given would be:
\[
5100 \cdot 499 \frac{5}{100} \times \frac{4}{99}
\]
However, since "5100" and "5100⋅499" are not standard notation for expressing the probabilities and don't quite fit here directly, the clearest answer to select based on the interpretation of choosing two rare cards without replacement is:
\[
\frac{5}{100} \times \frac{4}{99}
\]