A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
I am helpless regarding this. I don't know how to solve it. My teacher asked me to solve it by finding the probability that the balls drawn are blue and then subtracting it from 1. But I want to solve it straight forward and directly. Is it possible? If yes, how?
 A: Assumption: The first ball drawn is not put back in the bag.
Let E be the event that none of the two picked balls are blue. Further let $E_1$ and $E_2$ be the events that the first ball picked is not blue and the second ball picked is not blue respectively.
$P(E) = P(E_1 \cap E_2) = P(E_1)P(E_2 | E_1)$
$\implies P(E) = \left( {5 \choose 1}*1/7 \right) \left( {4 \choose 1}*1/6 \right)$
$\implies P(E) =  10/21 $
A: Let $X$ be the random variable for the number of blue drawn balls.
Then 
$P(X=0)=\frac{5}{7}\cdot \frac{4}{6}=\frac{10}{21}$

My teacher asked me to solve it by finding the probability that the
  balls drawn are blue and then subtracting it from 1.

To calulate $P(X=0)$ this is not right, because 
$P(X=0)+P(X=1)+P(X=2)=1$
Solving for $P(X=0)$
$P(X=0)=1-P(X=1)-P(X=2)$
$=1-2\cdot \frac{2}{7}\cdot \frac{5}{6}-\frac{2}{7}\cdot \frac{1}{6}=\frac{10}{21}$
Note that at $X=1$ you have two cases:


*

*First ball blue, second ball red or green.

*First ball red or green, second ball blue.
That´s why the middle summand has the additional factor $2$.
A: You can also calculate the number of possible pairs of balls (assuming the ball is drawn and not thrown back in), which is $\binom{7}{2} = 21$ . Now calculate the number of possible pairs given only red and green ball. This gives us $\binom{5}{2}= 10$ 
So the probability is $\frac{\text{number of red and green ball pairs}}{\text{number of all possible pairs} }=\frac{10}{21}$
A: Use multiplication rule to find the probability of not drawing the blue twice assuming without replacement.
\begin{equation}
P(\textit{draw red or green twice}) = \frac{5}{7} \cdot \frac{4}{6} = \frac{20}{42} = \frac{10}{21}. 
\end{equation}
A: In a bag there are 4 red balls ,2green balls, and 3yellow balls .if two balls are drawn at random ,what is the probability that at least one of the balls is green in color.
