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In a bag there are $4$ white, $4$ red and $2$ green balls. Two balls are selected at random. What is the probability that at least one ball is of green color?

Attempt- Total ways are clear i.e. $^{10}C_2=45$.

Now for at least two green balls we add cases of 1 green ball with 1 other color ball and both green balls.

For first case (1 green ball with 1 other color ball)- $^2C_1 \times ^8C_1=2\times 8=16 $

Second case (both green)= $^2C_2=1$

Therefore, Probabiliy should be $17/45$, but my book gives answer $18/45=2/5$, which clearly means that the book is calculating 2 ways to pick 2 green balls from 2 green balls.

I am confused here? We are not arranging balls here, just picking, so it should be $^2C_2=1$, right?

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  • $\begingroup$ You are correct. The probability that no green ball is selected is $\frac8{10}\frac79=\frac{28}{45}=1-\frac{17}{45}$. $\endgroup$ – drhab May 4 '17 at 13:27
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I get that the number of ways to pick both non-green balls is $^8C_2=28$, so you are right, the book is wrong!

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