Question - Impossibility of Successive Even Number Picks
Solution:
The image displays three cards with the numbers 7, 8, and 9 on them and asks for the probability of picking an even number and then picking an even number again.However, given these three cards, this task isn't actually possible, because there is only one even number (8) among them. Once you pick the even number 8, there are only two odd numbers left (7 and 9), which means you can't pick another even number since no other even numbers are available.For illustrative purposes, let's calculate the probability as if these were the only cards and you are picking without replacement:1. Probability of picking an even number (the 8) on the first draw:$$P(\text{even first}) = \frac{\text{Number of even cards}}{\text{Total number of cards}} = \frac{1}{3}$$2. Since this is without replacement, we now have two cards left, both of which are odd. The probability of picking another even card from the remaining two cards is:$$P(\text{even second} | \text{even first}) = \frac{\text{Number of remaining even cards}}{\text{Total remaining cards}} = \frac{0}{2}$$Since it is not possible to select another even card after the first even card has been selected, the overall probability of picking an even number and then another even number is zero. The probability of an impossible event is always 0.
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