Example Question - picking cards
Here are examples of questions we've helped users solve.
Finding Probability of Picking Even Numbers from a Set of Cards
This problem is about finding the probability of picking an even number from a set of cards labeled with numbers from 1 to 7, and then, without replacing the first card, picking another even number. Let's determine the probability step by step. 1. The probability of picking an even number (2, 4, 6) on the first draw: There are 3 even numbers out of 7 total numbers, so the probability is \( \frac{3}{7} \). 2. The probability of picking another even number on the second draw: After one even card is removed, there are now 2 even numbers remaining, and only 6 cards in total to choose from. So the probability for the second draw is \( \frac{2}{6} \) which simplifies to \( \frac{1}{3} \). Now we multiply the probabilities of both events happening in sequence, which are independent in this context: \( \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} \), which simplifies to \( \frac{1}{7} \). So, the probability of drawing an even number and then another even number without replacement is \( \frac{1}{7} \).
Impossible Event: Picking Even Numbers
The image shows the numbers 7, 8, and 9, and we are asked to calculate the probability of picking an even number and then picking another even number. Firstly, let's understand the scenario. We have three numbers: 7, 8, and 9. Out of these, only one number is even: 8. Since we are not replacing the first card before picking the second card, the probabilities will be different for each draw. The probability of picking the even number (which is 8) on the first draw is 1 out of 3, since there is only one even number out of three possibilities. This can be written as: \[ P(\text{first draw even}) = \frac{1}{3} \] After picking an even number on the first draw, there are now only two cards left, but since we've already picked the only even card, there are no even cards left. This means that the probability of picking an even number now is 0, as there are no even numbers to choose from. Therefore, the probability of picking an even number second time after already having picked an even number first is: \[ P(\text{second draw even} | \text{first draw even}) = 0 \] To find the combined probability of both events happening in sequence (first picking an even number, then picking another even number), we multiply the two individual probabilities together: \[ P(\text{first even and second even}) = P(\text{first draw even}) \times P(\text{second draw even} | \text{first draw even}) \] \[ P(\text{first even and second even}) = \frac{1}{3} \times 0 = 0 \] Thus, the probability of picking an even number and then picking another even number is 0, or simply impossible in this scenario.
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