Since order is not important in the selection and that we are to select 6 from the 25 transistors, the transistors may be selected by "combination of 25 taken 6 or 25C6 ways" which is equal to 177100. For the given condition, the transistor must selected from 19 which are not defective. There are 19C6 ways of doing so which is equal to 27132. Dividing this value by 177100 gives an answer of 0.15.