Try this solution:
1. for probability of both televisions work:
[tex] P_{2w}=\frac{C^2_8}{C^2_{10}}=\frac{7*8}{9*10}=0.6(2) [/tex]
2. for probability of at least on is defective (it means, the sum = one is defective and two are defective):
[tex] P_{1d}=\frac{C^1_2C^1_8}{C^2_{10}}=\frac{4*8}{9*10}=0.3(5) [/tex] - for P(1defective);
[tex] P_{2d}=\frac{C^2_2}{C^2_{10}}=\frac{2}{9*10}=0.0(2) [/tex] - for P(2defective);
totaly for at least one is defective:
[tex] P_{at \ least \ 1 \ d}=P_{1d}+P_{2d}=\frac{16}{45} +\frac{1}{45} =0.3(7) [/tex]
