Correct option is C. 17/19

The product of 3 randomly chosen integers are even only if there are even nos.

P(even) = 1 – P(odd)

P(odd) = \(\cfrac{10}{20}\times\cfrac9{19}\times\cfrac8{18}\)

= \(\cfrac2{19}\)

P(even) = \(\cfrac{17}{19}\)

3 integers out of 20 can be chosen in ²⁰C₃ ways. 

As in first 20 integers, 10 are even. 

To have the product of 3 integer even the chosen integers must be even.

∴ 3 integers that are even can be chosen from first 20 integers in ¹⁰C₃ ways. 

∴ Probability =  \(\frac{^{10}C_2}{^{20}C_2}\) = \(\frac{10\times9\times8}{20\times19\times18} = \frac{2}{19}\) 

Our answer matches with option (a) 

∴ Option (a) is the only correct choice

Correct Answer – C

Correct Answer – B
N/a