There are 6 tickets to the theater, four of which are for seats in the front row. 3 tickets are selected at random. What is the probability that two of them are for the front row?

Out of all the 6 tickets, 4 are front row tickets and the rest 2 are not.

Out of 4 front tickets 2 can be selected in $$^{4\ }C\ _2$$

The remaining 1 ticket has to be selected in  $$^{2\ }C\ _1$$

Hence total ways of selecting 3 tickets such that 2 of them are front row are  $$^{4\ }C\ _2$$ * $$^{2\ }C\ _1$$

3 out of 6 tickets can be selected in $$^{6\ }C\ _3$$ ways.

Required probability = $$^{4\ }C\ _2$$ * $$^{2\ }C\ _1$$ / $$^{6\ }C\ _3$$

Which is equal to 0.6