Question

you pick a card at random. without putting the first card back, you pick a second card at random.4 5 6 7what is the probability of picking an even number and then picking an even number?simplify your answer and write it as a fraction or whole number.

Explanation:

Step1: Calculate the probability of picking an even - numbered card first

There are 4 cards in total (\(n = 4\)), and the even - numbered cards are 4 and 6 (\(m_1=2\)). The probability formula is \(P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So the probability of picking an even - numbered card first, \(P_1=\frac{2}{4}=\frac{1}{2}\).

Step2: Calculate the probability of picking an even - numbered card second

After picking an even - numbered card first (without replacement), there are \(n_2 = 3\) cards left. If the first card was even, there is \(m_2 = 1\) even - numbered card left. So the probability of picking an even - numbered card second, \(P_2=\frac{1}{3}\).

Step3: Calculate the probability of the combined event

Since these are two dependent events, the probability of both events occurring is \(P = P_1\times P_2\). Substitute \(P_1=\frac{1}{2}\) and \(P_2=\frac{1}{3}\) into the formula: \(P=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\).

Answer:

\(\frac{1}{6}\)