**(c)** \(\frac{16}{33}\)

Here two articles are chosen together without replacement

P(one of the articles is substandard) = P (first article is standard) × P(second article is substandard) + P (first article is substandard) × P (second article is standard)

= \(\frac{60}{100}\times\frac{40}{99}\) + \(\frac{40}{100}\times\frac{60}{99}\)

= \(\frac{4800}{9900}\) = \(\frac{48}{99}\) = \(\frac{16}{33}\)