Question

you get a card at random. without putting the first card back, you get a second card at random.

what 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 number first

There are 3 cards: 5, 6, 7. The even - numbered card is 6. So the probability of picking an even number first, \(P_1=\frac{1}{3}\).

Step2: Calculate the probability of picking an even number second (without replacement)

After picking one card (the even - numbered 6), there are 2 cards left. Since the first even - numbered card is already picked, the number of favorable outcomes (picking an even number) for the second draw is 0. So the probability of picking an even number second, \(P_2 = 0\).

Step3: Use the multiplication rule for independent (in the sense of sequential without replacement) events

The multiplication rule for two events \(A\) and \(B\) is \(P(A\cap B)=P(A)\times P(B|A)\). Here, \(A\) is "picking an even number first" and \(B\) is "picking an even number second". So \(P = \frac{1}{3}\times0\).

Answer:

\(0\)