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 probability of first - pick

There are 3 even numbers (4, 6, 8) out of 6 cards. So the probability of picking an even number first, $P_1=\frac{3}{6}=\frac{1}{2}$.

Step2: Calculate probability of second - pick

After picking an even - numbered card first (without replacement), there are 5 cards left. There are 3 odd numbers (5, 7, 9). So the probability of picking an odd number second, given that an even number was picked first, $P_2=\frac{3}{5}$.

Step3: Calculate combined probability

Since these are dependent events, the probability of picking an even number first and then an odd number is $P = P_1\times P_2$. Substitute $P_1=\frac{1}{2}$ and $P_2=\frac{3}{5}$ into the formula: $P=\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}$.

Answer:

$\frac{3}{10}$