Verified answer

25 winning tickets, 475 losing tickets.

Since b and c specify "exactly",

we'll take a to mean "one or more".

P(one or more wins) = 1 - P(zero)

P(zero) = 475 * 474 * 473 * 472 * 471 * 471 / (500 * 499 * 498 * 497 * 496 * 495) = 0.735

Hence P(one or more) = 0.265 or 26.5%

Exactly 2 could be:

W W L L L L =

25 * 24 * 475 * 474 * 473 * 472 / (500 * 499 * 498 * 497 * 496 * 495) = 0.001989

But the 2 winning tickets could be in

any of 6c2 = 15 pairs of positions among the 6 tickets,

so P(exactly 2) = 15 * the above = 0.0298, or roughly 3 %

Exactly 4 is calculated similarly:

25 * 24 * 23 * 22 * 475 * 474 * 6c4 / (500 * 499 * 498 * 497 * 496 * 495) =

0.0000676

or 0%

This is comparable to the simpler, but less accurate formula:

15 * .05^4 * .95^2 = 0.0000846