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 8 and then picking an 8? write your answer as a fraction or whole number.

Explanation:

Step1: Calculate first - pick probability

There are 3 cards in total, and 1 of them is an 8. The probability of picking an 8 on the first pick is $\frac{1}{3}$.

Step2: Calculate second - pick probability

After picking an 8 on the first pick without replacement, there are 2 cards left and 0 cards with 8. So the probability of picking an 8 on the second pick given that an 8 was picked on the first pick is 0.

Step3: Calculate combined probability

Since the two events are sequential and we use the multiplication rule for independent - like events in a non - replacement situation $P(A\cap B)=P(A)\times P(B|A)$. Here $P(A)=\frac{1}{3}$ and $P(B|A) = 0$, so the probability of picking an 8 and then an 8 is $\frac{1}{3}\times0 = 0$.

Answer:

0