A bag contains 7 red checkers and 11 black checkers. One checker is drawn and replaced. Then a second checker is drawn. What is the probability that a red checker was drawn both times?

there are 18 checkers in the bag each time and each time 7 of them are red

first time 7/18
second time 7/18
49/324 or about 0.15

Note:

This question is really the same as the four aces question you asked below.

To find the probability that a red checker was drawn both times, we need to calculate the probability of drawing a red checker on the first draw and then drawing a red checker again on the second draw.

Let's break it down step by step:

Step 1: Calculate the probability of drawing a red checker on the first draw.
There are a total of 7 red checkers and 11 black checkers, so the probability of drawing a red checker on the first draw is 7/(7+11) = 7/18.

Step 2: Since we replaced the first checker back into the bag before the second draw, the probabilities remain the same for the second draw.
So, the probability of drawing a red checker on the second draw is also 7/18.

Step 3: Multiply the probabilities from step 1 and step 2 together.
The required probability is the product of the probabilities from step 1 and step 2, which is (7/18) * (7/18).

So, the probability of drawing a red checker both times is (7/18) * (7/18) = 49/324.

Therefore, the probability that a red checker was drawn both times is 49/324.