How to find the different possibilities in picking 4 balls from bag of 3Red 2 blue balls with repetition?

[naveen]

I have 3 Red balls and 2 blue balls in bag.

If I pick 4 balls from bag continuously , replacing the ball picked into the bag.

The possible ways getting one red and 3 blue balls are:
1R 1B 1B 1B
1B 1R 1B 1B
1B 1B 1R 1B
1B 1B 1B 1R

there are 4 possibilities, how to derive this without having to plot the balls manually ?

 

[Dr.Peterson]

I have 3 Red balls and 2 blue balls in bag.

If I pick 4 balls from bag continuously , replacing the ball picked into the bag.

The possible ways getting one red and 3 blue balls are:
1R 1B 1B 1B
1B 1R 1B 1B
1B 1B 1R 1B
1B 1B 1B 1R

there are 4 possibilities, how to derive this without having to plot the balls manually ?

I suppose "continuously" means "sequentially", one after another. The important word is "replacing".

We can list the four ways more compactly as RBBB, BRBB, BBRB, and BBBR. We can think of these as if we were given a set of four spaces, _ _ _ _, and want to put an R into one of them, and B into the rest. How many ways are there to select one of four places? This is a combination problem: \(_4C_1 = \frac{4!}{1!3!} = 4\).

You can also think of this as permutation the "word" RBBB; you may learn a formula for permutations like this at some point.

I sense a question about probability coming; that hasn't been asked or answered yet.