You pick a card at random. Without putting the first card back, you pick a second card at random.678What is the probability of picking a prime number and then picking a prime number?Simplify your answer and write it as a fraction or whole number.
Question
You pick a card at random. Without putting the first card back, you pick a second card at random.678What is the probability of picking a prime number and then picking a prime number?Simplify your answer and write it as a fraction or whole number.
Solution
The question seems to be about picking cards with prime numbers, but it doesn't specify the range of numbers on the cards. Prime numbers are numbers that have only two distinct positive divisors: 1 and the number itself. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, etc.
Let's assume that the cards are numbered from 1 to 100. There are 25 prime numbers between 1 and 100.
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The probability of picking a prime number on the first draw is the number of favorable outcomes (prime numbers) divided by the total number of outcomes (total numbers). So, the probability is 25/100 = 1/4.
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Since you're not replacing the first card, there are now 99 cards left. If the first card was a prime number, there are now 24 prime numbers left. So, the probability of picking a prime number on the second draw is 24/99.
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The probability of both events happening is the product of their individual probabilities. So, the probability of drawing a prime number twice in a row is (1/4) * (24/99) = 24/396 = 1/16.5.
Please note that this answer is based on the assumption that the cards are numbered from 1 to 100. If the range of numbers is different, the answer would also be different.
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