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 an even number and then picking an odd 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 8 cards in total. The even - numbered cards are 2, 4, 6, 8. So the number of even - numbered cards \(n(E_1)=4\).
The probability of picking an even - numbered card first \(P(E_1)=\frac{4}{8}=\frac{1}{2}\).

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

After picking an even - numbered card first (without replacement), there are 7 cards left. The odd - numbered cards are 1, 3, 5, 7. So the number of odd - numbered cards \(n(O_2) = 4\).
The probability of picking an odd - numbered card second given that an even - numbered card was picked first \(P(O_2|E_1)=\frac{4}{7}\).

Step3: Use the multiplication rule for dependent events

The multiplication rule for two dependent events \(A\) and \(B\) is \(P(A\cap B)=P(A)\times P(B|A)\).
Let \(A\) be the event of picking an even - numbered card first and \(B\) be the event of picking an odd - numbered card second.
So \(P(E_1\cap O_2)=P(E_1)\times P(O_2|E_1)=\frac{1}{2}\times\frac{4}{7}\).

$$P(E_1\cap O_2)=\frac{4}{14}=\frac{2}{7}$$

Answer:

\(\frac{2}{7}\)