ColdplayMX

The total number of balls is 23. The probability of choosing a blue ball on her first draw is 3/23. The probabilty of choosing a red ball on her second draw is 12/22, or 6/11, because the ball is not replaced. Now we can multiply the probabilities and get \(3\over 4\)*\(6\over11\), equaling 18/44, or 9/22.

a 1. sorry. :(

Graphing it helps a lot.

An element is randomly chosen from among the first 15 rows. I might have left that out.

Simply calculate the values of the combinations and multiply for the first problem.

The second problem is really just 13578^-1, which you can calculate.

Try solving.

Please help someone!!!

Guest, if you want an answer you should probably include the data table in your problem description.

There is no data table in your problem description.