Pen Probability 9F08F6

1. **State the problem:** We have a pencil case with 8 blue pens and 4 green pens, total $8 + 4 = 12$ pens. 2. **Understanding the problem:** We pick one pen at random without replacement, then pick a second pen. We want the probability that the first pen is green and the second pen is blue. 3. **Set up the probabilities for the first pick:** - Probability of first pen green: $\frac{4}{12} = \frac{1}{3}$ - Probability of first pen blue: $\frac{8}{12} = \frac{2}{3}$ 4. **Set up the probabilities for the second pick given the first pick:** - If first pen is green (1 green taken), remaining pens: 11 total, blue still 8. Probability second pen blue given first green: $\frac{8}{11}$ - If first pen is blue (1 blue taken), remaining pens: 11 total, green still 4. Probability second pen green given first blue: $\frac{4}{11}$ 5. **Complete the tree diagram probabilities:** - First pen green $\frac{1}{3}$, then second pen blue $\frac{8}{11}$ - First pen green $\frac{1}{3}$, then second pen green $\frac{3}{11}$ - First pen blue $\frac{2}{3}$, then second pen blue $\frac{7}{11}$ - First pen blue $\frac{2}{3}$, then second pen green $\frac{4}{11}$ 6. **Calculate the required probability:** Probability(first green and second blue) = Probability(first green) $\times$ Probability(second blue given first green) $$ = \frac{1}{3} \times \frac{8}{11} = \frac{8}{33} $$ **Final answer:** The probability that the first pen is green and the second pen is blue is $\boxed{\frac{8}{33}}$.