A box contains $3$ blue, $4$ red, $5$ green balls. Six balls are taken out at random and without replacement. What is the probability exactly four are red, at least one is green?
I noticed that it might be $4$R, $1$G and $1$ random or $4$R, $2$G, but I just can't solve it.
On
There are total (3b + 4r + 5g)=12 balls in the box. Six balls are chosen. The no of ways of picking exactly 4 red balls from the total 4 red balls is 4C4. Similarly for at least 1 green ball, there are 5C1 + 5C2 + 5C3 + 5C4 + 5C5 ways. But the total balls to picked is 6, so we consider 5C1 + 5C2 only
Now,4 red balls are compulsory, if we take 1 green ball, we can choose 1 blue ball. This is possible in 3C1 ways.
if we take 2 green balls,there is no way to choose a blue ball.
So favourable outcomes: 4r1g1b and 4r2g = (4C4 * 5C1 * 3C1) + (4C4 * 5C2) ways
There are 2 possible cases: $$4R,2G \hspace{0.2cm} and \hspace{0.2cm} 4R,1G,1B$$
Total balls are $$4R,5G \hspace{0.2cm} and \hspace{0.2cm} 3B$$ For $4R,2G$: select balls in $$^4C_4 \cdot ^5C_2$$ For $4R,1G,1B$: select balls in $$^4C_4\cdot ^5C_1\cdot ^3C_1$$ Total ways to select 6 balls is $$^{4+5+3}C_6=^{12}C_6$$ Just divide:
$$P(4R,2G)=\frac{^4C_4\cdot ^5C_2}{^{12}C_6}$$ $$P(4R,1G,1B)=\frac{^4C_4\cdot ^5C_1\cdot ^3C_1}{^{12}C_6}$$
For 4R and at least 1G, add these two probabilities: $$ \frac{^4C_4 \cdot ^5C_2+ ^4C_4 \cdot ^5C_1\cdot ^3C_1}{^{12}C_6}$$