gooner wrote:
I see that using \(P(at least 1 red)\) = \(1 - P(no red)\) is the quick way to solve this.
However, should \(P(at least 1 red) = P(1 red) + P(2 red)\) work?
where \(P(2 red) = \frac{3C2}{5C2}\) and \(P(1 red)= \frac{2(3C1*2C1)}{5C2}\)
Its not working ofcource because \(P(1 red)= \frac{2(3C1*2C1)}{5C2}\) is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)
Where am I wrong in my thinking?
Thanks for the help
Hi Gooner,
Here order of choosing Blue and Red do not matter. Hence do not multiply with 2.
\(P(1 red)= \frac{(3C1*2C1)}{5C2}\)
\(P(2 red)= \frac{(3C2)}{5C2}\)
Finally probability of choosing atleast one red is P(1 red) + p(2 red) = \(\frac{(3C1*2C1)}{5C2}\) + \(\frac{(3C2)}{5C2}\).
i.e., 3+6/10 = 9/10.