two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability that jane will be on the same committee as albert ?
total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability = ( 8 / 20 ) * 100 = 40 % . ans : : d
multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4))