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I am trying to improve my stat skills from a book (Freedman et al, 2007, Statistics) that gives an exercise, I cannot get my head around.

It goes like this:

There are 5 green balls and 1 red ball in a box. We draw four times randomly with replacement. What is the probability that we draw at least two red balls?

My guess would be

(1/6)^2 + (1/6)^3 + (1/6)^4 because the probability of at least 2 red balls out of 4 draws must be equal to drawing two or three or four times a red ball. Hence I would use the addition rule, which gives 0.03 = 3 percent

However, the book says it is 13 percent, and gives the following explanation (150+20+1)/1296.

Why is this so?

My attempt to trace back the terms:

I can possibly see how it got the first and third term in the fraction: The first could be (5/6)^2 = 25/36, which is the probability of drawing two green balls; and the third could be (1/6)^4 = 1/1296, which is the probability of drawing four red balls. I have no idea though where they could possibly get the 20 from.

Thanks

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    The logic is correct. The terms are wrong. See Mark's answer. – 2017-01-16

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The probability of getting exactly 2 red balls is $$ \left(\frac16\right)^2\left(\frac56\right)^2\binom{4}{2} = \frac{25}{1296}\cdot 6 $$ where the binomial coefficient is just the number of ways you can select which two picks, out of the four, got red balls.

The probability of getting exactly 3 red balls is $$ \left(\frac16\right)^3\left(\frac56\right)^1\binom{4}{1} = \frac{5}{1296}\cdot 4 $$ where the binomial coefficient is just the number of ways you can select which one pick, out of the four, got a red ball. That is where the $20$ comes from.

Can you take it from here?

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    Yes, the probablity of getting exactly 4 red balls is: (1/6)^4 x 4!/(4! 0!) = 1/1296 I forgot the binomial formula, thank you! – 2017-01-16