To find the probability of getting exactly one new pen, we consider the two cases: one new pen and one old pen.
The probability of selecting one new pen first and then one old pen is:
\(P(N, \overline{N}) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} = \frac{7}{30}\)
The probability of selecting one old pen first and then one new pen is:
\(P(\overline{N}, N) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90} = \frac{7}{30}\)
Adding these probabilities gives:
\(P(\text{exactly one new pen}) = \frac{7}{30} + \frac{7}{30} = \frac{14}{30} = \frac{7}{15}\)
Alternatively, using combinations:
Total ways to choose 2 pens from 10 is \(\binom{10}{2} = 45\).
Ways to choose 1 new pen from 3 and 1 old pen from 7 is \(\binom{3}{1} \times \binom{7}{1} = 3 \times 7 = 21\).
Thus, the probability is:
\(\frac{21}{45} = \frac{7}{15}\)
