you pick a card at random. without putting the first card back, you pick a second card at random. what is…

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? write your answer as a percentage.

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? write your answer as a percentage.

Answer

Explanation:

Step1: Calculate probability of first - pick

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

Step2: Calculate probability of second - pick

After picking an even number first (without replacement), there are 5 numbers left, and 3 odd numbers (5, 7, 9). So the probability of picking an odd number second, $P_2 = \frac{3}{5}$.

Step3: Calculate combined probability

Since these are independent - like (in the sense of sequential) events, the probability of both events occurring is the product of their probabilities. $P=P_1\times P_2=\frac{3}{6}\times\frac{3}{5}=\frac{9}{30}=\frac{3}{10}$.

Step4: Convert to percentage

To convert $\frac{3}{10}$ to a percentage, we multiply by 100. $P = \frac{3}{10}\times100%=30%$.

Answer:

30%