A company recruits employees and specifies three examination courses. There are two examination schemes. Option 1: pass at least two of the three courses;

We assume that the passing events of the three examination courses are A, B and C respectively, and the corresponding probabilities are A, B and C..

(1) At least two of the three courses have passed, which can be expressed as AB. BC+ABC+ABC。 Let its probability

P 1, then p1= ab (1-c)+a (1-b) c+(1-a) BC+ABC = ab+AC+BC-2abc.

Set up three courses, choose two courses at random, and the probability of passing both courses is P2.

Then p2 =13ab+13ac+13bc.

(2)p 1-P2 =(a b+ AC+BC-2 ABC)-( 13ab+ 13ac+ 13bc)= 23ab+23ac+23bc-2 ABC

=23(ab+ac+bc-3abc)

= 23〔ab( 1-c)+AC( 1-b)+BC( 1-a)〕> 0

∴ p1> the probability of P2 using the first scheme is greater than the probability of using the second scheme.