Question
What is the probability of drawing an even numbered card, keeping it, and then drawing another even numbered card from a stack of 24 cards numbered 1 to 24?
Answer options with 5 options
A.
StartFraction 1 over 23 EndFraction
B.
StartFraction 2 over 23 EndFraction
C.
StartFraction 1 over 12 EndFraction
D.
StartFraction 1 over 11 EndFraction
E.
StartFraction 11 over 46 EndFraction
Answer options with 5 options
A.
StartFraction 1 over 23 EndFraction
B.
StartFraction 2 over 23 EndFraction
C.
StartFraction 1 over 12 EndFraction
D.
StartFraction 1 over 11 EndFraction
E.
StartFraction 11 over 46 EndFraction
Answers
GPT-4o mini
To calculate the probability of drawing an even numbered card from a stack of 24 cards numbered 1 to 24, keeping it, and then drawing another even numbered card, we can follow these steps:
1. **Count the even numbered cards**: In the range of 1 to 24, the even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, and 24. There are a total of 12 even cards.
2. **Calculate the probability of drawing the first even card**: When we draw the first card, there are 24 cards total, and 12 of them are even. So, the probability of drawing an even card first is:
\[
P(\text{first even}) = \frac{12}{24} = \frac{1}{2}
\]
3. **After drawing one even card**: After drawing and keeping one even card, there are now 23 cards remaining in the stack (24 - 1 = 23), and there will be 11 even cards left (12 - 1 = 11).
4. **Calculate the probability of drawing a second even card**: The probability of then drawing another even card from the remaining 23 cards is:
\[
P(\text{second even} | \text{first even}) = \frac{11}{23}
\]
5. **Combine the probabilities**: The overall probability of both events (drawing the first even card and then drawing a second even card) is:
\[
P(\text{first even and second even}) = P(\text{first even}) \times P(\text{second even} | \text{first even}) = \frac{12}{24} \times \frac{11}{23} = \frac{1}{2} \times \frac{11}{23} = \frac{11}{46}
\]
Thus, the final answer, which is the probability of drawing an even numbered card, keeping it, and then drawing another even numbered card from the stack, is:
**E.** \( \frac{11}{46} \)
1. **Count the even numbered cards**: In the range of 1 to 24, the even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, and 24. There are a total of 12 even cards.
2. **Calculate the probability of drawing the first even card**: When we draw the first card, there are 24 cards total, and 12 of them are even. So, the probability of drawing an even card first is:
\[
P(\text{first even}) = \frac{12}{24} = \frac{1}{2}
\]
3. **After drawing one even card**: After drawing and keeping one even card, there are now 23 cards remaining in the stack (24 - 1 = 23), and there will be 11 even cards left (12 - 1 = 11).
4. **Calculate the probability of drawing a second even card**: The probability of then drawing another even card from the remaining 23 cards is:
\[
P(\text{second even} | \text{first even}) = \frac{11}{23}
\]
5. **Combine the probabilities**: The overall probability of both events (drawing the first even card and then drawing a second even card) is:
\[
P(\text{first even and second even}) = P(\text{first even}) \times P(\text{second even} | \text{first even}) = \frac{12}{24} \times \frac{11}{23} = \frac{1}{2} \times \frac{11}{23} = \frac{11}{46}
\]
Thus, the final answer, which is the probability of drawing an even numbered card, keeping it, and then drawing another even numbered card from the stack, is:
**E.** \( \frac{11}{46} \)