Explanation

(a)
Out of 90 tickets, two tickets already considered, instead of selecting 5 tickets we have to select only 3 tickets out of 88 tickets.
$\therefore$ Required probability $= \frac{{\ ^{88}C}_{3}}{{\ ^{90}C}_{5}}$
$= \frac{\frac{88 \times 87 \times 86}{3 \times 2 \times 1}}{\frac{90 \times 89 \times 88 \times 87 \times 86}{5 \times 4 \times 3 \times 2 \times 1}}$
$= \frac{5 \times 4}{90 \times 89} = \frac{2}{801}$