Question

you pick a card at random. without putting the first card back, you pick a second card at random. what is the probability of picking a prime number and then picking a prime number? write your answer as a percentage rounded to the nearest tenth. %

Explanation:

Step1: Determine the number of prime numbers

The prime numbers in the set \(\{1,2,3,4,5,6,7,8\}\) are \(2\), \(3\), \(5\), \(7\). So there are \(n = 4\) prime numbers out of \(N=8\) cards initially.
The probability of picking a prime number on the first draw, \(P_1=\frac{4}{8}\)

Step2: Calculate the probability for the second draw

After picking one prime - numbered card (without replacement), there are \(n_1 = 3\) prime numbers left and \(N_1=7\) cards left.
The probability of picking a prime number on the second draw given a prime number was picked on the first draw, \(P_2=\frac{3}{7}\)

Step3: Use the multiplication rule for dependent events

The probability of both events (picking a prime number first and then another prime number) is \(P = P_1\times P_2\)

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Step4: Convert to a percentage

To convert the decimal to a percentage, we multiply by \(100\). So \(P = 0.2143\times100\%\approx21.4\%\)

Answer:

\(21.4\%\)