Probability of Successive Even Number Selections without Replacement
The image shows three cards with the numbers 7, 8, and 9. To find the probability of picking an even number and then picking an even number again (without replacing the first card), we follow these steps:
1. Determine the probability of picking an even number on the first draw. There is only one even number (8), and there are three cards total, so the probability is \( \frac{1}{3} \).
2. After picking an even number the first time, there are now only two cards left, and since we've taken out the 8 (the only even number), there are no even numbers left. Therefore, the probability of picking an even number on the second draw is \( \frac{0}{2} \) because there are zero even numbers out of the two remaining cards.
The probability of both independent events happening is the product of their individual probabilities:
\( \frac{1}{3} \times \frac{0}{2} = 0 \)
Thus, the probability of picking an even number and then picking an even number again without replacement is 0. This makes sense intuitively, because once we've picked the only even number (8), it's not possible to pick another even number since it won't be replaced.
