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. save answer

Explanation:

Step1: Count prime numbers

Prime numbers in \(2,3,4,5\) are \(2,3,5\). So there are \(3\) prime numbers out of \(4\) cards initially.
The probability of picking a prime number first is \(P_1=\frac{3}{4}\).

Step2: Calculate probability for the second pick

After picking one prime number (without replacement), there are \(2\) prime numbers left and \(3\) cards in total.
The probability of picking a prime number second is \(P_2 = \frac{2}{3}\).

Step3: Use multiplication rule for dependent events

The probability of both events (picking prime then prime) is \(P=P_1\times P_2\).

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Step4: Convert to percentage

To convert \(\frac{1}{2}\) to a percentage, we use the formula \(P(\%)=\frac{1}{2}\times100\%\).
\(P(\%) = 50.0\%\)

Answer:

\(50.0\%\)