Question

A sample of 4 different calculators is randomly selected from a group

Answer 1

$\displaystyle\\|\Omega|=\binom{72}{4}=\dfrac{72!}{4!68!}=\dfrac{69\cdot70\cdot71\cdot72}{2\cdot3\cdot4}=1028790\\|A|=\binom{46}{4}=\dfrac{46!}{4!42!}=\dfrac{43\cdot44\cdot45\cdot46}{2\cdot3\cdot4}=163185\\\\P(A)=\dfrac{163185}{1028790}=\dfrac{473}{2982}\approx0.1586$