Example Question - picking numbers
Here are examples of questions we've helped users solve.
Calculating Probability of Picking Numbers Without Replacement
To solve this problem, we need to determine the probability of picking a 6 first and then a 5 without replacing the first number. The total number of possible numbers to choose from initially is 8 (since we see numbers 2 through 9). The probability of picking a 6 first is: 1 out of 8 (since there is only one 6) After picking a 6, there are now 7 numbers left to choose from. The probability of picking a 5 next is: 1 out of 7 (since there is only one 5) To find the combined probability of both events occurring (picking a 6 and then a 5), we multiply the probabilities of each individual event: \( (1/8) * (1/7) = 1/56 \) As a percentage, \( (1/56) \) can be calculated as follows: \( (1/56) * 100 \approx 1.786\% \) Rounding to the nearest tenth, the answer is 1.8%.
Probability of Picking a Specific Sequence of Numbers from a Set of Cards
The image shows a set of cards with numbers from 2 through 9. The question asks for the probability of picking a 6 and then picking a 5. To solve this problem, we need to calculate the probability of picking a 6 first, and then, given that a 6 has been picked, the probability of then picking a 5. The total number of cards is 8 (2 through 9). The probability of picking a 6 first is 1 out of 8, because there is only one 6 in the set of cards. After picking a 6, there are 7 cards left. The probability of picking a 5 from the remaining cards is 1 out of 7, because there is only one 5 in the remaining set of cards. The probability of both events happening in sequence (picking a 6 and then picking a 5) is the product of the probabilities of each individual event happening: Probability of picking a 6 and then a 5 = (Probability of picking a 6) × (Probability of picking a 5 given a 6 was picked) = (1/8) × (1/7) = 1/56 So the probability of picking a 6 and then picking a 5 is 1/56.
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