Three integers are chosen from 1 to 20 without replacement. Probability that their product is even, is 

Step-I:  

$\text{ Out of }20\text{ integers, }3\text{ can be selected in }{}^{20}C_3\text{ ways}$

$\mathrm{to} \mathrm{be} \mathrm{product} \mathrm{even}$at least one integer is even

= Total-odd  

$={20}_{C_3}-{10}_{C_3}$

$R·P=\frac{{20}_{C_3}-{10}_{C_3}}{{20}_{C_3}}=1-\frac{{10}_{C_3}}{{20}_{C_3}}$

$=\frac{17}{19}=0.89$